Two problems force calculus into existence: finding the slope of a tangent line, and finding instantaneous velocity. This deck starts there, then gives the informal definition of a limit and shows how to estimate limits from tables and read them off graphs. It covers one-sided limits and the two-sided existence test, removable holes, the three ways a limit can fail to exist, and infinite limits and vertical asymptotes, before closing with an honest first look at epsilon-delta. It targets the classic traps: assuming the limit is the function value, declaring that a limit exists when the one-sided limits disagree, calling an infinite limit an existing limit, and trusting a table on an oscillating function.
Subject: Calculus I · 132 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 01
What it means for a function to head somewhere, and how to see it in a table, on a graph, and right at a hole.
Objectives
This deck builds the single idea that everything else in Calculus I rests on. By the end you can:
Section
Section 1
Concept
Slope is easy on a straight line. Pick any two points and take rise over run.
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
On a curve the steepness changes at every point. Asking for the slope at one single point breaks the formula: one point gives you no rise and no run, only the useless fraction below.
\[ \frac{0}{0} \]
tangent line — The straight line that touches a curve at one point and matches the curve's direction there. Its slope is what we mean by 'the slope of the curve at that point'.
Counterexample
Discussion prompt
Slope is easy on a straight line. Pick any two points and take rise over run.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Picture it
Animation
Shows: The problem that started calculus — a rendered Manim animation.
Rendered with Manim.
Takeaway: How do you find the slope of a curve at a single point?
Intuition
Put a smooth curve under a microscope. Turn the magnification up, and up again. A small enough piece of a smooth curve is indistinguishable from a straight line.
That straight line you see at maximum zoom is the tangent line. The curve has a slope at a point because, up close, it is a line.
The trouble is that you never actually arrive at infinite zoom. You only ever get closer. A limit is the tool that lets you talk about where that zooming is headed without ever finishing it.
Analogy
Discussion prompt
Explain Zoom in until the curve looks straight by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Put a smooth curve under a microscope. Turn the magnification up, and up again. A small enough piece of a smooth curve is indistinguishable from a straight line.
Concept
Figure (svg): A parabola with a point P at (1,1); a secant line drawn from P to a second point Q at (2,4), and the tangent line at P shown separately.
secant line — A line through two points of a curve. Its slope is an honest rise-over-run, because two points really do give a rise and a run.
Here is the trick. Keep one point fixed and slide the second point along the curve toward it. Each position gives a secant slope you can actually compute.
As the second point closes in, the secant lines pivot and settle onto the tangent. The tangent slope is the number those secant slopes are heading for.
Definition probe
Sort into buckets
Every line below is part of the definition of tangent line or of secant line — one or the other, never both. Put each where it belongs.
Ranking
Put in order
Put the moves of Secant slopes closing in on a tangent slope into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Anything on the parabola has the form of an input paired with its square, so shifting the input by h forces the output to be the square of the shifted input.
Worked example
Find the slope of the tangent to this parabola at the marked point.
\[ y = x^2, \qquad P = (1,\, 1) \]
Put the second point a distance h away from P along the curve
Why: Anything on the parabola has the form of an input paired with its square, so shifting the input by h forces the output to be the square of the shifted input.
\[ Q = \left(1 + h,\ (1+h)^2\right) \]
Write the secant slope through P and Q, then simplify
Why: Rise over run for two real points is always legal. Expanding the square lets the h in the denominator cancel, which is the whole point of the algebra.
\[ m_{\text{sec}} = \frac{(1+h)^2 - 1}{h} = \frac{1 + 2h + h^2 - 1}{h} = \frac{2h + h^2}{h} = 2 + h \]
Shrink h toward zero from both sides and watch the number
Why: We may not set h equal to zero, because the original fraction would be zero over zero. But we may make h as small as we like and see where the slopes are going.
| h | second point Q | secant slope |
|---|---|---|
| -0.5 | (0.5, 0.25) | 1.5 |
| -0.1 | (0.9, 0.81) | 1.9 |
| -0.01 | (0.99, 0.9801) | 1.99 |
| 0.01 | (1.01, 1.0201) | 2.01 |
| 0.1 | (1.1, 1.21) | 2.1 |
| 0.5 | (1.5, 2.25) | 2.5 |
State the tangent slope
Why: From both sides the secant slopes squeeze in on the same number, and the simplified expression makes that obvious.
\[ m_{\text{tan}} = 2 \]
Verify by checking the tangent line touches the curve at P
Why: The line through the point with slope 2 is y = 2x - 1. At the input 1 it gives 1, which is exactly the height of the parabola there, so the line really does touch at P. And at h = 0.01 the secant slope 2.01 sits within one hundredth of 2, as predicted.
\[ y = 2x - 1 \quad \Rightarrow \quad y(1) = 2(1) - 1 = 1 = 1^2 \ \checkmark \]
Concept
Average velocity is easy: distance travelled divided by the time it took. That is a rise over a run in disguise.
\[ v_{\text{avg}} = \frac{\Delta s}{\Delta t} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \]
But your speedometer reads a velocity right now, over no elapsed time at all. Put the same instant in both slots and the formula collapses again.
So we do the same dodge as before: compute the average over a short interval, then make the interval shorter and shorter and see what number the averages approach.
Picture it
Animation
Shows: Average speed, then instantaneous speed — a rendered Manim animation.
Rendered with Manim.
Takeaway: The odometer gives averages. The speedometer takes the limit.
Step zero
Discussion prompt
How fast is the rock falling at one second? — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write the average velocity from one second to one second plus h
Answer:
Worked example
A rock is dropped off a cliff. Ignoring air resistance, the distance fallen in feet after t seconds is given below. Find its velocity at the instant one second after release.
\[ s(t) = 16t^2 \]
Write the average velocity from one second to one second plus h
Why: This is an interval of real length, so the ordinary average-velocity formula is perfectly legal here.
\[ v_{\text{avg}} = \frac{16(1+h)^2 - 16(1)^2}{h} \]
Expand and cancel the h
Why: Expanding the square produces a common factor of h in the numerator, and cancelling it removes the zero-over-zero problem so we can see the behavior.
\[ \frac{16\left(1 + 2h + h^2\right) - 16}{h} = \frac{32h + 16h^2}{h} = 32 + 16h \]
Tabulate the averages over shorter and shorter intervals
Why: Each row is a real measurement you could make with a stopwatch. The pattern in the last column is the evidence.
| interval (s) | h | average velocity (ft/s) |
|---|---|---|
| 1 to 2 | 1 | 48 |
| 1 to 1.5 | 0.5 | 40 |
| 1 to 1.1 | 0.1 | 33.6 |
| 1 to 1.01 | 0.01 | 32.16 |
| 1 to 1.001 | 0.001 | 32.016 |
Report the instantaneous velocity
Why: As the interval shrinks the averages march toward a single value, and the simplified form 32 plus 16h shows it exactly.
\[ v(1) = 32 \ \text{ft/s} \]
Verify with an interval on the other side of one second
Why: Approaching from below must give the same number. Using the interval from 0.99 to 1 second, h is negative one hundredth, and the average velocity computes to 31.84 ft/s, which matches 32 plus 16 times negative one hundredth. Both sides close in on 32.
\[ \frac{16(0.99)^2 - 16}{-0.01} = \frac{15.6816 - 16}{-0.01} = 31.84 = 32 + 16(-0.01) \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "How fast is the rock falling at one second?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Approaching from below must give the same number. Using the interval from 0.99 to 1 second, h is negative one hundredth, and the average velocity computes to 31.84 ft/s, which matches 32 plus 16 times negative one hundredth. Both sides close in on 32.
Concept
The tangent problem and the velocity problem look unrelated. They are the same problem.
| problem | what you can compute | what you actually want |
|---|---|---|
| tangent slope | secant slope over a gap | the slope as the gap closes |
| instant velocity | average velocity over an interval | the average as the interval closes |
In both cases the quantity you want is unreachable by direct computation, but the quantities you can compute line up and point at it.
A limit is the machinery for saying what they point at. Learn limits once and both problems, plus every rate-of-change problem after them, are solved.
Comparison
Comparison matrix
From Two problems, one question underneath: refill the what you can compute column from what you know. The rest of the table is as it appeared.
| problem | what you can compute | what you actually want |
|---|---|---|
| tangent slope | secant slope over a gap | the slope as the gap closes |
| instant velocity | average velocity over an interval | the average as the interval closes |
Section
Section 2
Concept
limit — The number L is the limit of f(x) as x approaches a if the outputs f(x) get and stay arbitrarily close to L whenever x is close enough to a, but not equal to a.
\[ \lim_{x \to a} f(x) = L \]
Three phrases in that definition are load-bearing. Arbitrarily close means as close as anyone demands. Close enough means you get to decide how near to squeeze. But not equal means the input value itself is excluded on purpose.
Intuition
Imagine driving toward a town on a straight road. At every mile marker you can read the sign and see where you are headed. You do not have to arrive to know the destination.
A limit reads the signs. It answers: where is this function headed as the input closes in? Whether the function ever gets there, or what happens if it does, is a different question.
This is why a limit can exist at a point where the function is not even defined. The road can be closed at the town line and the signs still say where the road was going.
Explain it
Discussion prompt
Explain The destination, not the arrival to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Imagine driving toward a town on a straight road. At every mile marker you can read the sign and see where you are headed. You do not have to arrive to know the destination.
Concept
\[ \lim_{x \to 3} \left(x^2 + 1\right) = 10 \]
Say it as: the limit, as x approaches 3, of x squared plus one, equals ten. The arrow is read approaches, never equals.
Two writing habits worth forming now. The word lim is never left dangling without its arrow underneath, and the equals sign only appears once you have finished taking the limit.
| piece | what it means |
|---|---|
| the arrow expression underneath | which input the variable is closing in on |
| the expression after lim | the function whose outputs you are watching |
| the number after the equals sign | where those outputs are headed |
Trade off
Comparison matrix
From Reading the notation out loud: every row here is a choice with a cost. Fill the what it means column, then say which row you would actually pick and what you give up for it.
| piece | what it means |
|---|---|
| the arrow expression underneath | which input the variable is closing in on |
| the expression after lim | the function whose outputs you are watching |
| the number after the equals sign | where those outputs are headed |
Picture it
Animation
Shows: Reading the notation out loud — a rendered Manim animation.
Rendered with Manim.
Takeaway: Say it in words once and the symbols stop being decoration.
Concept
Nothing about a limit at a point depends on what the function does far away. Change the function out at a thousand and the limit near three is untouched.
All that matters is a small neighborhood of inputs on either side of the target, with the target itself punched out.
\[ 0 < |x - a| < \delta \]
That double inequality is the whole idea in symbols: within a small distance of the target, and strictly more than zero away from it. We will come back to it at the end of the deck.
Picture it
Animation
Shows: A limit is a local question — a rendered Manim animation.
Rendered with Manim.
Takeaway: Only the neighbourhood of the point matters. The far ends are irrelevant.
Concept
This is the single most misunderstood fact in the whole topic, so read it twice.
The definition explicitly excludes the target input. The function may be undefined there, or defined with a wildly wrong value, and the limit is unaffected.
\[ \lim_{x \to a} f(x) \ \text{ and } \ f(a) \ \text{ are two different questions} \]
When they happen to agree, the function is called continuous at that point, and that is a special extra property, not the default.
Picture it
Animation
Shows: The algebra of cancelling a shared factor, leaving a hole at the point.
Heading, not arriving.
Takeaway: A limit asks where the function is heading, never where it lands. The value at the point can be wrong, or missing entirely, without changing the answer.
Intuition
Picture a smooth road with one square of pavement removed. Standing on either side, you can see exactly what height the missing square should be.
The limit is that height: the value the road is heading toward from both directions. The hole does not change it.
Now imagine someone drops a traffic cone in the hole at the wrong height. The road still heads toward the same place. The cone is the function value, and it is simply a different fact from the limit.
Estimation
Predict first
All three of these functions have the same limit as the input approaches two, and they disagree about the point itself.
Commit before you compute: what does Three functions, one limit come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify all three with the same nearby inputs
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At an input of 1.99 all three functions output 4.99, and at 2.01 all three output 5.01.
Worked example
All three of these functions have the same limit as the input approaches two, and they disagree about the point itself.
\[ f(x) = x + 3 \]
Read off the first one: it is a line, so the outputs march straight to the obvious height
Why: A line has no gaps, so the value you are heading toward is the value you land on. Here the limit and the function value agree.
\[ \lim_{x \to 2} f(x) = 5, \qquad f(2) = 5 \]
Factor the second one and cancel
Why: The numerator factors, and the cancelled factor is nonzero for every input except two, which the limit excludes anyway. So on the region the limit looks at, the second function IS the line.
\[ g(x) = \frac{x^2 + x - 6}{x - 2} = \frac{(x+3)(x-2)}{x-2} = x + 3 \quad (x \neq 2) \]
State the second limit and the second function value
Why: The limit is the same 5, but plugging two into the original gives zero over zero, so the function has no value there at all.
\[ \lim_{x \to 2} g(x) = 5, \qquad g(2) \ \text{is undefined} \]
Handle the third one, which has a value deliberately placed at the wrong height
Why: Everywhere except the single point two the third function agrees with the line, and the limit only looks everywhere except that point.
\[ h(x) = \begin{cases} x + 3, & x \neq 2 \\ 1, & x = 2 \end{cases} \qquad \lim_{x \to 2} h(x) = 5, \ h(2) = 1 \]
Verify all three with the same nearby inputs
Why: At an input of 1.99 all three functions output 4.99, and at 2.01 all three output 5.01. Since the limit only inspects inputs like these, the three limits must be identical, and they are all 5. The disagreement lives entirely at the single excluded point.
| x | f(x) | g(x) | h(x) |
|---|---|---|---|
| 1.99 | 4.99 | 4.99 | 4.99 |
| 2 | 5 | undefined | 1 |
| 2.01 | 5.01 | 5.01 | 5.01 |
Picture it
Animation
Shows: Each line of the worked example "Three functions, one limit", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At an input of 1.99 all three functions output 4.99, and at 2.01 all three output 5.01. Since the limit only inspects inputs like these, the three limits must be identical, and they are all 5. The disagreement lives entirely at the single excluded point.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The instinct from algebra is to plug the number in and report whatever comes out.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Substituting gives zero over zero, so the student writes 'undefined, the limit does not exist' and stops.
Ask instead what the outputs do at inputs near two, never at two itself.
Why: Substituting gives zero over zero, so the student writes 'undefined, the limit does not exist' and stops.
Trap
The instinct from algebra is to plug the number in and report whatever comes out.
\[ \lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} \]
Substitute two directly and declare the limit undefined
Why: Substituting gives zero over zero, so the student writes 'undefined, the limit does not exist' and stops.
\[ \frac{2^2 + 2 - 6}{2 - 2} = \frac{0}{0} \ \Rightarrow \ \text{``limit does not exist''} \ \times \]
The error, named
Why: Zero over zero is not an answer. It is a signal that direct substitution has failed and the limit question is still wide open. The function value at two was never what the limit asked about.
Ask instead what the outputs do at inputs near two, never at two itself.
\[ \lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} \]
Factor and cancel the offending factor, then substitute
Why: For every input except two the fraction is literally equal to the simpler expression, and the limit ignores the input two by definition.
\[ \frac{(x+3)(x-2)}{x-2} = x+3 \ (x \neq 2) \ \Rightarrow \ \lim_{x \to 2} \frac{x^2+x-6}{x-2} = 5 \ \checkmark \]
Say both facts out loud
Why: The limit is 5 and the function value at two does not exist. Both statements are true at the same time, and a graph of this function is the line with one point punched out.
Notation
Annotate
From Trap: assuming the limit equals the function value — read this one piece at a time. What is each part doing?
On: \( \lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} \)
Section
Section 3
Concept
A limit table is a numerical stakeout. You choose inputs that creep toward the target and record what the function does.
Both sides matter because a function can head for two different heights depending on which way you come in. A one-sided table would hide that completely.
Picture it
Animation
Shows: A table of inputs approaching 2 with outputs approaching 4.
Evidence, not proof.
Takeaway: A table suggests an answer and builds confidence, but it never proves one — no finite list of points rules out a surprise closer in.
Step zero
Discussion prompt
Estimating a limit numerically — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Notice direct substitution fails
Answer:
Worked example
Estimate this limit from a table, then confirm it algebraically.
\[ \lim_{x \to 1} \frac{x^2 - 1}{x - 1} \]
Notice direct substitution fails
Why: Putting one into numerator and denominator gives zero over zero, which is a signal to investigate, not an answer. A table is the fastest first look.
\[ \frac{1^2 - 1}{1 - 1} = \frac{0}{0} \quad \text{(no information yet)} \]
Compute the function at inputs closing in from the left
Why: Each of these inputs is a legal input, since none of them is one. For example at 0.99 the numerator is negative 0.0199 and the denominator is negative 0.01, giving 1.99.
| x (from the left) | value of the quotient |
|---|---|
| 0.9 | 1.9 |
| 0.99 | 1.99 |
| 0.999 | 1.999 |
Do the same from the right
Why: At 1.01 the numerator is 0.0201 and the denominator is 0.01, giving 2.01. The two sides are closing in on the same place from opposite directions.
| x (from the right) | value of the quotient |
|---|---|
| 1.001 | 2.001 |
| 1.01 | 2.01 |
| 1.1 | 2.1 |
Read the estimate off the table
Why: Both columns are squeezing toward the same number, and every extra decimal of closeness buys an extra decimal of agreement.
\[ \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2 \]
Verify the estimate by factoring the numerator
Why: The numerator is a difference of squares, so the troublesome factor cancels for every input other than one. What is left is a line, and its height at one is exactly the 2 the table predicted.
\[ \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1 \ (x \neq 1) \ \Rightarrow \ 1 + 1 = 2 \ \checkmark \]
Worked example
Estimate this limit numerically. Substitution will fail again, and this time the answer is not a whole number.
\[ \lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} \]
Check substitution first so you know a table is needed
Why: The square root of four is two, so the numerator is zero, and the denominator is zero as well. Direct substitution gives no information.
\[ \frac{\sqrt{0+4}-2}{0} = \frac{0}{0} \]
Tabulate both sides, keeping six decimal places
Why: The values here differ in the fourth decimal place, so a short table with two decimals would be useless. Extra precision is what makes the pattern visible.
| x | value of the quotient |
|---|---|
| -0.1 | 0.251582 |
| -0.01 | 0.250156 |
| -0.001 | 0.250016 |
| 0.001 | 0.249984 |
| 0.01 | 0.249844 |
| 0.1 | 0.248457 |
Read the estimate
Why: The left column is dropping toward a value and the right column is rising toward the same value, and both are pinned near one quarter.
\[ \lim_{x \to 0} \frac{\sqrt{x+4}-2}{x} = \frac{1}{4} = 0.25 \]
Verify by multiplying by the conjugate
Why: Multiplying top and bottom by the conjugate of the numerator turns the numerator into a difference of squares, the x cancels, and what remains is a continuous expression you can substitute into directly. It gives one over four, matching the table to every digit shown.
\[ \frac{\sqrt{x+4}-2}{x} \cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} = \frac{x}{x\left(\sqrt{x+4}+2\right)} = \frac{1}{\sqrt{x+4}+2} \ \longrightarrow \ \frac{1}{4} \ \checkmark \]
Picture it
Animation
Shows: When the table lands on a fraction — a rendered Manim animation.
Rendered with Manim.
Takeaway: Decimals suggest the fraction. Algebra has to confirm it.
Concept
A table is evidence, not proof. It shows you a handful of inputs out of infinitely many.
| a table is good at | a table can lie about |
|---|---|
| suggesting the value quickly | functions that wiggle between your sample points |
| revealing a two-sided disagreement | answers that need more digits than you kept |
| sanity-checking algebra you already did | round-off in a calculator near a cancellation |
So the working rule is: use a table to form a conjecture, then confirm it with algebra or with a graph. Never hand in a table as the entire argument.
Comparison
Comparison matrix
From What a table can and cannot prove: refill the a table can lie about column from what you know. The rest of the table is as it appeared.
| a table is good at | a table can lie about |
|---|---|
| suggesting the value quickly | functions that wiggle between your sample points |
| revealing a two-sided disagreement | answers that need more digits than you kept |
| sanity-checking algebra you already did | round-off in a calculator near a cancellation |
Pattern
Whenever you are asked to estimate a limit numerically, run these five moves in order.
Step five is the one students skip, and it is the one that catches oscillating functions before they embarrass you.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A student tabulates this function at neat, convenient inputs approaching zero.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The chosen inputs are all reciprocals of whole numbers, so the angle is always a whole multiple of half a turn, where the sine is exactly zero.
Before trusting a table, ask whether the function could be wiggling between your sample points. Then deliberately sample inputs that break the pattern.
Why: The chosen inputs are all reciprocals of whole numbers, so the angle is always a whole multiple of half a turn, where the sine is exactly zero. The table looks airtight.
Trap
A student tabulates this function at neat, convenient inputs approaching zero.
\[ f(x) = \sin\!\left(\frac{\pi}{x}\right), \qquad \lim_{x \to 0} f(x) = ? \]
Every single row reads zero, so the student concludes the limit is zero
Why: The chosen inputs are all reciprocals of whole numbers, so the angle is always a whole multiple of half a turn, where the sine is exactly zero. The table looks airtight.
| x | value of the sine |
|---|---|
| 1 | 0 |
| 0.5 | 0 |
| 0.1 | 0 |
| 0.01 | 0 |
| 0.001 | 0 |
The error, named
Why: The table sampled only the inputs that happen to give zero. Between any two of those rows the function has already swung all the way to one and all the way to negative one. The conclusion is wrong.
Before trusting a table, ask whether the function could be wiggling between your sample points. Then deliberately sample inputs that break the pattern.
\[ f(x) = \sin\!\left(\frac{\pi}{x}\right) \]
Pick inputs that land on the peaks and troughs instead
Why: Choosing inputs of the form two over an odd number makes the angle an odd multiple of a quarter turn, where the sine is exactly one or negative one. These inputs are just as close to zero as the ones in the bad table.
| x | angle | value of the sine |
|---|---|---|
| 0.4 | 2.5 half-turns | 1 |
| 0.285714... | 3.5 half-turns | -1 |
| 0.222222... | 4.5 half-turns | 1 |
| 0.181818... | 5.5 half-turns | -1 |
Conclude correctly
Why: Arbitrarily close to zero the function takes the value one AND the value negative one, over and over forever. The outputs never settle, so the limit does not exist.
\[ \lim_{x \to 0} \sin\!\left(\frac{\pi}{x}\right) \ \text{does not exist} \]
Invariant
Step through it
Step through Trap: trusting a table too far one row at a time. One of these columns never changes — find it, and say why it cannot.
Elimination
Eliminate the wrong options
What is the limit of this quotient as x approaches 2?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Both columns close in on 12 from opposite sides. Factoring confirms it: the difference of cubes gives (x - 2) times (x squared plus 2x plus 4), the (x - 2) cancels, and substituting 2 into what remains gives 4 plus 4 plus 4, which is 12.
Check
A student is estimating the limit below and produces this table. Decide what the table is telling you before you look at the choices.
\[ \lim_{x \to 2} \frac{x^3 - 8}{x - 2} \]
| x | value of the quotient |
|---|---|
| 1.9 | 11.41 |
| 1.99 | 11.9401 |
| 1.999 | 11.994001 |
| 2.001 | 12.006001 |
| 2.01 | 12.0601 |
| 2.1 | 12.61 |
Check your understanding
What is the limit of this quotient as x approaches 2?
Answer: A
Why: Both columns close in on 12 from opposite sides. Factoring confirms it: the difference of cubes gives (x - 2) times (x squared plus 2x plus 4), the (x - 2) cancels, and substituting 2 into what remains gives 4 plus 4 plus 4, which is 12.
Pattern
Step through it
Step through Check yourself: what does the table suggest? one row at a time. What is driving the change, and what would the row after the last one be?
Section
Section 4
Concept
To read a limit off a picture, put your finger on the curve to the left of the target input and slide it in. Watch the height your finger is heading for, not where it stops.
Do the same from the right. If both approaches head for the same height, that height is the limit.
Open circles and filled dots are the graph's way of telling you about the point itself, and the sliding finger never reaches them. Read them separately.
| symbol on the graph | what it tells you |
|---|---|
| open circle | the function is not defined at that height there |
| filled dot | that is the actual function value there |
| nothing marked | read the value straight off the curve |
Trade off
Comparison matrix
From A limit on a graph is a height you are heading for: every row here is a choice with a cost. Fill the what it tells you column, then say which row you would actually pick and what you give up for it.
| symbol on the graph | what it tells you |
|---|---|
| open circle | the function is not defined at that height there |
| filled dot | that is the actual function value there |
| nothing marked | read the value straight off the curve |
Estimation
Predict first
The graph beside you is a straight climb, then a straight descent, with a piece missing at the peak and a stray dot placed lower down.
Commit before you compute: what does Reading a limit and a value off the same picture come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the graph reading with two nearby numbers
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At an input of 1.9 the climbing branch gives 2.9, and at an input of 2.1 the descending branch gives 2.9 as well.
Worked example
Figure (svg): A tent-shaped graph rising to a peak at x equals 2 with an open circle at height 3, and a filled dot at height 1 directly below it.
The graph beside you is a straight climb, then a straight descent, with a piece missing at the peak and a stray dot placed lower down.
\[ f(x) = \begin{cases} x + 1, & x < 2 \\ 1, & x = 2 \\ 5 - x, & x > 2 \end{cases} \]
Slide in from the left and read the height you approach
Why: The climbing branch is the line one more than the input, and just to the left of two that height is just under three. The open circle sits at three.
\[ \lim_{x \to 2^{-}} f(x) = 3 \]
Slide in from the right and read that height
Why: The descending branch is five minus the input, and just to the right of two that height is just under three as well. Both approaches aim at the same open circle.
\[ \lim_{x \to 2^{+}} f(x) = 3 \]
Combine the two readings into the two-sided limit
Why: The approaches agree, so the two-sided limit exists and equals their common value. Nothing about the filled dot enters this reasoning.
\[ \lim_{x \to 2} f(x) = 3 \]
Read the function value separately
Why: The filled dot is the graph telling you the actual output at two. It is a different question from the limit, and here the two answers disagree.
\[ f(2) = 1 \neq 3 = \lim_{x \to 2} f(x) \]
Verify the graph reading with two nearby numbers
Why: At an input of 1.9 the climbing branch gives 2.9, and at an input of 2.1 the descending branch gives 2.9 as well. Both are within a tenth of 3, exactly as the picture promised.
| x | which branch | f(x) |
|---|---|---|
| 1.9 | x plus 1 | 2.9 |
| 1.99 | x plus 1 | 2.99 |
| 2 | the stray dot | 1 |
| 2.01 | 5 minus x | 2.99 |
| 2.1 | 5 minus x | 2.9 |
Concept
Sometimes the two approaches disagree. To even describe that situation we need language for each direction separately.
left-hand limit — The value the outputs approach as the input closes in on the target from below, using only inputs less than the target.
right-hand limit — The value the outputs approach as the input closes in on the target from above, using only inputs greater than the target.
Picture it
Animation
Shows: A step-shaped graph whose left and right limits differ at the origin.
Both sides exist. They just do not match.
Takeaway: Each one-sided limit can exist perfectly well on its own. If they disagree, the two-sided limit simply does not exist.
Intuition
Two people walk down a hallway toward the same doorway, one from the east and one from the west, calling out the room number they see as they go.
If they are shouting the same number as they close in, the room has one clear identity: that is the two-sided limit.
If one is shouting three and the other is shouting seven, there is no single answer to give. The function is doing two different things at the same spot, and the two-sided limit simply does not exist.
Concept
\[ \lim_{x \to a^{-}} f(x) \qquad \text{and} \qquad \lim_{x \to a^{+}} f(x) \]
The little raised minus means from the left, from inputs smaller than the target. The little raised plus means from the right, from inputs larger than the target.
Read the marks as directions on the number line, not as signs of the number itself. Approaching negative four from the left is still marked with a raised minus.
\[ \lim_{x \to -4^{-}} f(x) \quad \text{means inputs like } -4.1,\, -4.01,\, -4.001 \]
Picture it
Animation
Shows: Notation for the two directions — a rendered Manim animation.
Rendered with Manim.
Takeaway: The little sign is the whole difference.
Ranking
Put in order
Put the moves of One-sided limits at a seam into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The left-hand limit looks exclusively at inputs less than one, and on that whole region the function is the squaring formula.
Worked example
This function is built from two different formulas that meet at the input one. Find both one-sided limits there and the function value.
\[ g(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \ge 1 \end{cases} \]
For the left-hand limit, use only the formula that governs inputs below one
Why: The left-hand limit looks exclusively at inputs less than one, and on that whole region the function is the squaring formula. Which formula is in force at the seam itself is irrelevant.
\[ \lim_{x \to 1^{-}} g(x) = \lim_{x \to 1^{-}} x^2 = 1 \]
For the right-hand limit, use only the formula that governs inputs above one
Why: Inputs greater than one are all handled by the linear piece, so the right-hand limit is that line's height as the input closes in on one.
\[ \lim_{x \to 1^{+}} g(x) = \lim_{x \to 1^{+}} (2x+1) = 3 \]
Compare the two and report the two-sided limit
Why: The two directions head for different heights, one and three. There is no single number the outputs settle on, so the two-sided limit fails to exist. This is a jump discontinuity.
\[ 1 \neq 3 \ \Rightarrow \ \lim_{x \to 1} g(x) \ \text{does not exist} \]
Read the function value, which is a separate question again
Why: The seam belongs to the linear piece because that piece includes the endpoint, so the output at one comes from the linear formula.
\[ g(1) = 2(1) + 1 = 3 \]
Verify with numbers just either side of the seam
Why: At 0.999 the squaring formula gives 0.998001, and at 1.001 the linear formula gives 3.002. Those two are nowhere near each other even though the inputs differ by two thousandths, which is exactly what a genuine jump looks like numerically.
| x | formula in force | g(x) |
|---|---|---|
| 0.99 | x squared | 0.9801 |
| 0.999 | x squared | 0.998001 |
| 1 | 2x plus 1 | 3 |
| 1.001 | 2x plus 1 | 3.002 |
| 1.01 | 2x plus 1 | 3.02 |
Concept
This is the theorem that makes one-sided limits worth defining, and it is an exactly when statement, working in both directions.
\[ \lim_{x \to a} f(x) = L \iff \lim_{x \to a^{-}} f(x) = L \ \text{ and } \ \lim_{x \to a^{+}} f(x) = L \]
In words: the two-sided limit exists exactly when both one-sided limits exist and are equal, and then it is their common value.
That gives you a decision procedure. Compute both sides. Same finite number, and you have a limit. Anything else, and you do not.
Picture it
Animation
Shows: A smooth curve with a point approached from the left and from the right.
Two approaches, one destination.
Takeaway: Walk in from the left, walk in from the right. The limit exists precisely when the two arrivals agree on a height.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A student computes both one-sided limits correctly for the piecewise function from before, then tries to force out a single answer anyway.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It feels reasonable: the function is heading for one from below and three from above, so surely two splits the difference.
Apply the existence test as written. It is a yes-or-no question, and the answer here is no.
Why: It feels reasonable: the function is heading for one from below and three from above, so surely two splits the difference. The student writes that the limit equals two.
Trap
A student computes both one-sided limits correctly for the piecewise function from before, then tries to force out a single answer anyway.
\[ \lim_{x \to 1^{-}} g(x) = 1, \qquad \lim_{x \to 1^{+}} g(x) = 3 \]
Average the two one-sided values and report the middle
Why: It feels reasonable: the function is heading for one from below and three from above, so surely two splits the difference. The student writes that the limit equals two.
\[ \frac{1+3}{2} = 2 \quad \Rightarrow \quad \text{``}\lim_{x \to 1} g(x) = 2\text{''} \ \times \]
The error, named
Why: The function never takes any value near two anywhere close to the seam. Check it: at 0.999 the output is 0.998 and at 1.001 the output is 3.002. Nothing is near two, so two cannot be a value the outputs are getting arbitrarily close to.
| x | g(x) | distance from 2 |
|---|---|---|
| 0.999 | 0.998001 | about 1.002 |
| 1.001 | 3.002 | about 1.002 |
Apply the existence test as written. It is a yes-or-no question, and the answer here is no.
\[ \lim_{x \to 1^{-}} g(x) = 1, \qquad \lim_{x \to 1^{+}} g(x) = 3 \]
Compare, then declare
Why: The two one-sided limits are unequal, so by the existence test there is no number that the outputs approach from both directions. The correct answer is that the limit does not exist.
\[ 1 \neq 3 \ \Rightarrow \ \lim_{x \to 1} g(x) \ \text{does not exist} \ \checkmark \]
Then say what IS true, because 'does not exist' is not the whole story
Why: A good answer names both one-sided limits and identifies the behavior as a jump of size two. That is far more informative than a bare statement that something failed.
| question | answer |
|---|---|
| left-hand limit | 1 |
| right-hand limit | 3 |
| two-sided limit | does not exist (jump) |
| function value | 3 |
Discrimination
Sort into buckets
Sort these by answer, from memory, without looking back at Trap: one-sided limits that disagree. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Ranking
Put in order
These are the steps of Pattern: read any limit off a graph, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Given a picture and a target input, run this every time.
The thumb is not a joke. Physically hiding the point is the fastest cure for the habit of reading the dot instead of the approach.
Check
Figure (svg): A straight line rising from the origin with an open circle at the point where x equals 1 and height 2, and a filled dot directly above it at height 4.
Cover the point with your thumb, trace in from each side, and decide before you read the choices.
Check your understanding
From the graph, what is the two-sided limit of this function as x approaches 1?
Answer: A
Why: Tracing in from the left the height climbs toward the open circle at 2, and tracing in from the right it drops toward the same open circle at 2. Both one-sided limits are 2, so the two-sided limit is 2. The filled dot at 4 is the function value, which is a separate question.
Prediction
Predict first
What can you conclude about the two-sided limit of h as x approaches 5?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: It does not exist.
Why: The existence test says the two-sided limit exists exactly when both one-sided limits exist and are equal. Here they are negative 2 and 3, which are different, so no single number is approached from both sides and the two-sided limit does not exist.
Check
You are told these three facts about a function, and nothing else. Use the two-sided existence test.
\[ \lim_{x \to 5^{-}} h(x) = -2, \qquad \lim_{x \to 5^{+}} h(x) = 3, \qquad h(5) = -2 \]
Check your understanding
What can you conclude about the two-sided limit of h as x approaches 5?
Answer: A
Why: The existence test says the two-sided limit exists exactly when both one-sided limits exist and are equal. Here they are negative 2 and 3, which are different, so no single number is approached from both sides and the two-sided limit does not exist.
Section
Section 5
Concept
A limit fails to exist when the outputs refuse to settle on a single number. There are exactly three ways that happens, and each one looks different on a graph.
Notice what is not on the list: a hole. A hole removes a function value, and the limit was never asking about the function value.
Matching
Match the pairs
From Three ways a limit fails — match each one to what it actually does. The descriptions have been shuffled.
Why: Jump, Unbounded growth, Oscillation are easy to tell apart while they are sitting next to their descriptions and much harder afterwards, which is what this checks.
Concept
Figure (svg): A step graph: a flat segment at height 1 ending in an open circle at x equals 1, and a flat segment at height 3 starting with a filled dot at x equals 1.
A jump is a clean break: the graph is calm on both sides but lands at two different heights. Shipping costs, tax brackets, and parking rates all behave this way.
\[ \lim_{x \to 1^{-}} f(x) = 1, \qquad \lim_{x \to 1^{+}} f(x) = 3 \]
Both one-sided limits exist and are perfectly well behaved. They just disagree, and the existence test then says the two-sided limit does not exist.
Picture it
Animation
Shows: The three questions to ask before claiming a limit exists.
Answer all three, in order.
Takeaway: Ask whether each one-sided limit exists, then whether they agree. A jump answers yes, yes, no — and that last no is the whole failure.
Concept
Here the outputs do not disagree about a height. They refuse to have a height at all, running past every number you name.
\[ \lim_{x \to 0} \frac{1}{x^2} \]
| x | value |
|---|---|
| 0.1 | 100 |
| 0.01 | 10000 |
| 0.001 | 1000000 |
| 0.0001 | 100000000 |
No number L can be the limit, because the outputs eventually exceed L and never come back. We will give this failure its own special notation in a moment.
Pattern
Step through it
Step through Failure by unbounded growth one row at a time. What is driving the change, and what would the row after the last one be?
Picture it
Animation
Shows: A curve spiking upward near a point.
A description of the failure.
Takeaway: Growing without bound is not a limit. Writing infinity here describes how the limit failed; it does not name a number the function reached.
Concept
The third failure is the sneakiest. The outputs stay in a nice small range, so nothing runs off the page, but they never stop swinging.
\[ f(x) = \sin\!\left(\frac{1}{x}\right), \qquad x \to 0 \]
As the input shrinks toward zero, the quantity inside the sine grows without bound, so the sine races through complete cycles faster and faster. Near zero it completes infinitely many of them.
The outputs are always trapped between negative one and one, but they visit both ends forever. Being bounded is not the same as settling down.
Picture it
Animation
Shows: A curve oscillating faster and faster as it approaches zero.
Bounded, and still no limit.
Takeaway: The values stay bounded and still there is no limit — because they never settle. Boundedness is not enough; the function has to converge.
Hypothesis
Predict first
Proving an oscillating limit does not exist is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Find inputs that make the sine exactly zero
Why: The sine is zero at every whole multiple of half a turn, so choose the reciprocals of those angles. As the whole number grows these inputs shrink to zero.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Show that this limit does not exist. The strategy is to exhibit two families of inputs that both close in on zero but give different outputs forever.
\[ \lim_{x \to 0} \sin\!\left(\frac{1}{x}\right) \]
Find inputs that make the sine exactly zero
Why: The sine is zero at every whole multiple of half a turn, so choose the reciprocals of those angles. As the whole number grows these inputs shrink to zero.
\[ x_k = \frac{1}{k\pi} \ \Rightarrow \ \sin\!\left(\frac{1}{x_k}\right) = \sin(k\pi) = 0 \]
Find inputs that make the sine exactly one
Why: The sine equals one at a quarter turn plus any number of full turns. Taking reciprocals again gives inputs that also shrink to zero, interleaved with the first family.
\[ y_k = \frac{2}{(4k+1)\pi} \ \Rightarrow \ \sin\!\left(\frac{1}{y_k}\right) = \sin\!\left(\frac{\pi}{2} + 2k\pi\right) = 1 \]
Lay the two families side by side
Why: Both columns of inputs are marching to zero, and no matter how close to zero you look, both zeros and ones are still appearing. The outputs are not settling on anything.
| input giving 0 | input giving 1 |
|---|---|
| 0.318310 | 0.636620 |
| 0.159155 | 0.127324 |
| 0.106103 | 0.070736 |
| 0.031831 | 0.048970 |
Conclude
Why: If some number L were the limit, then eventually every output would be within one tenth of L. But outputs of 0 and outputs of 1 keep occurring arbitrarily close to zero, and no single L is within one tenth of both.
\[ \lim_{x \to 0} \sin\!\left(\frac{1}{x}\right) \ \text{does not exist} \]
Verify one entry from each column by hand
Why: Take the input two over nine times a half turn, about 0.070736. Its reciprocal is four and a half half-turns, and the sine of that is the sine of a quarter turn, which is exactly 1. Take the input one over ten half-turns, about 0.031831. Its reciprocal is ten half-turns, whose sine is exactly 0. Both inputs are smaller than one tenth, so both behaviors really do occur that close to zero.
\[ \sin\!\left(\frac{9\pi}{2}\right) = 1, \qquad \sin(10\pi) = 0 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Proving an oscillating limit does not exist", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Take the input two over nine times a half turn, about 0.070736. Its reciprocal is four and a half half-turns, and the sine of that is the sine of a quarter turn, which is exactly 1. Take the input one over ten half-turns, about 0.031831. Its reciprocal is ten half-turns, whose sine is exactly 0. Both inputs are smaller than one tenth, so both behaviors really do occur that close to zero.
Intuition
Think of a radio dial that spins faster the closer you get to the end of the band. Near the end it is spinning so fast that in any tiny stretch you sweep the whole range of stations.
There is no station you are tuning in to. Zooming in does not calm it down; zooming in makes it worse.
That is the real test for a limit: does zooming in make the outputs agree? Holes pass that test. Jumps fail it once. Oscillation fails it at every scale, forever.
Section
Section 6
Concept
\[ \lim_{x \to a} f(x) = \infty \]
Read this out loud as: the outputs grow without bound as the input approaches a. It is a description of a specific failure, written in the shape of an equation.
infinite limit — Notation stating that the outputs eventually exceed every number you name, and stay above it, as the input closes in on the target. It records how the limit fails; it does not name a number the function reaches.
The symbol on the right is not a number, so nothing on that line is an ordinary equation. Many textbooks say the limit does not exist but grows without bound, and that phrasing is the honest one.
Step zero
Discussion prompt
An infinite limit that agrees on both sides — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Check the denominator at the target
Answer:
Worked example
Analyze the behavior of this function near the input three.
\[ f(x) = \frac{1}{(x-3)^2} \]
Check the denominator at the target
Why: The denominator is zero exactly at three while the numerator is one, which is not zero. That combination is the signature of unbounded behavior rather than a hole.
\[ (3-3)^2 = 0, \qquad \text{numerator} = 1 \neq 0 \]
Decide the sign of the denominator on each side
Why: The denominator is a square, so it is positive for every input except three. Both approaches therefore divide a positive one by a tiny positive number.
\[ (x-3)^2 > 0 \quad \text{for all } x \neq 3 \]
Tabulate to see the scale of the growth
Why: Each time the input gets ten times closer, the denominator gets one hundred times smaller, so the output gets one hundred times larger. Nothing bounds it.
| x | value |
|---|---|
| 2.9 | 100 |
| 2.99 | 10000 |
| 2.999 | 1000000 |
| 3.001 | 1000000 |
| 3.01 | 10000 |
| 3.1 | 100 |
State the result in both directions and two-sided
Why: Because the sign is the same on both sides, the one-sided descriptions match and we may write the two-sided statement as well.
\[ \lim_{x \to 3^{-}} \frac{1}{(x-3)^2} = \infty, \quad \lim_{x \to 3^{+}} \frac{1}{(x-3)^2} = \infty, \quad \lim_{x \to 3} \frac{1}{(x-3)^2} = \infty \]
Verify by naming a threshold and beating it
Why: Suppose someone demands an output above one hundred million. Take the input 3.0001. Then the denominator is one hundred-millionth, and the output is exactly one hundred million, and any input closer than that gives more. Since this works for any demand, the growth really is unbounded, and the line at three is a vertical asymptote.
\[ f(3.0001) = \frac{1}{(0.0001)^2} = \frac{1}{10^{-8}} = 10^{8} \ \checkmark \]
Picture it
Animation
Shows: When both sides run away together — a rendered Manim animation.
Rendered with Manim.
Takeaway: Both sides head the same way, so the notation may say infinity.
Estimation
Predict first
Analyze this function near the input two. The denominator factors, and the two factors behave very differently there.
Commit before you compute: what does An infinite limit whose two sides disagree come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify one row by hand and state the conclusion
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At the input 2.01 the numerator is 1.01 and the denominator is 4.0401 minus 4, which is 0.0401.
Worked example
Analyze this function near the input two. The denominator factors, and the two factors behave very differently there.
\[ f(x) = \frac{x-1}{x^2-4} = \frac{x-1}{(x-2)(x+2)} \]
Confirm this is an asymptote and not a hole
Why: At the input two the numerator is one, which is not zero, and the denominator is zero. Nothing cancels, so the function blows up rather than filling in.
\[ \text{at } x = 2: \ \text{numerator} = 1, \ \text{denominator} = 0 \]
Track the sign of each factor from the right
Why: Just above two the factor x minus two is a tiny positive number, x plus two is about four, and the numerator is about one. Positive divided by tiny positive is enormously positive.
\[ x \to 2^{+}: \ \frac{(+1)}{(0^{+})(+4)} \longrightarrow \infty \]
Track the sign of each factor from the left
Why: Just below two the factor x minus two is a tiny negative number while the other two pieces keep their signs. Positive divided by tiny negative is enormously negative.
\[ x \to 2^{-}: \ \frac{(+1)}{(0^{-})(+4)} \longrightarrow -\infty \]
Confirm the sign analysis numerically
Why: The table shows the magnitudes exploding and the signs opposite, exactly as the factor analysis predicted.
| x | value (rounded) |
|---|---|
| 1.9 | -2.31 |
| 1.99 | -24.81 |
| 1.999 | -249.81 |
| 2.001 | 250.19 |
| 2.01 | 25.19 |
| 2.1 | 2.68 |
Verify one row by hand and state the conclusion
Why: At the input 2.01 the numerator is 1.01 and the denominator is 4.0401 minus 4, which is 0.0401. Dividing gives about 25.19, matching the table and confirming the positive sign from the right. Since one side runs up and the other runs down, there is no two-sided infinite statement to make: the two-sided limit simply does not exist, and the line at two is a vertical asymptote.
\[ \frac{2.01-1}{(2.01)^2-4} = \frac{1.01}{0.0401} \approx 25.19 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "An infinite limit whose two sides disagree", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the input 2.01 the numerator is 1.01 and the denominator is 4.0401 minus 4, which is 0.0401. Dividing gives about 25.19, matching the table and confirming the positive sign from the right. Since one side runs up and the other runs down, there is no two-sided infinite statement to make: the two-sided limit simply does not exist, and the line at two is a vertical asymptote.
Concept
Figure (svg): The graph of one over the quantity x minus two, with a dashed vertical line at x equals two; the left branch dives downward and the right branch shoots upward.
vertical asymptote — A vertical line such that at least one of the one-sided limits of the function there runs off to positive or negative unbounded values. The graph hugs the line without ever touching it.
Only one of the four possible one-sided behaviors is needed. The two sides do not have to agree, and often they do not, as the picture shows.
\[ \lim_{x \to 2^{-}} \frac{1}{x-2} = -\infty, \qquad \lim_{x \to 2^{+}} \frac{1}{x-2} = \infty \]
Concept
Both a hole and a vertical asymptote come from a denominator hitting zero. What separates them is what the numerator is doing at the same input.
| at the target input | numerator | denominator | behavior |
|---|---|---|---|
| case 1 | not zero | zero | vertical asymptote |
| case 2 | zero | zero | cancel first, then look again |
| case 3 | zero | not zero | the limit is zero, nothing special |
Case two is the interesting one. Factor and cancel, then re-examine. If the trouble cancels out completely you have a hole; if a factor of the denominator survives, you have an asymptote after all.
\[ \frac{x-4}{x^2-x-12} = \frac{x-4}{(x-4)(x+3)} = \frac{1}{x+3} \quad (x \neq 4) \]
That example has a hole at one of its two denominator zeros and a genuine asymptote at the other, which is exactly why you must factor before you answer.
Comparison
Comparison matrix
From Hole or asymptote? Look at the numerator: refill the numerator column from what you know. The rest of the table is as it appeared.
| at the target input | numerator | denominator | behavior |
|---|---|---|---|
| case 1 | not zero | zero | vertical asymptote |
| case 2 | zero | zero | cancel first, then look again |
| case 3 | zero | not zero | the limit is zero, nothing special |
Picture it
Animation
Shows: Hole or wall? Look at the numerator — a rendered Manim animation.
Rendered with Manim.
Takeaway: The denominator alone never decides it.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Asked whether the limit below exists, a student sees a clean equation written with an equals sign and answers yes.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The notation looks exactly like every other limit statement, so the student treats the right-hand side as an answer and writes that the limit exists.
Read the statement as a description of behavior, and answer the existence question separately.
Why: The notation looks exactly like every other limit statement, so the student treats the right-hand side as an answer and writes that the limit exists.
Trap
Asked whether the limit below exists, a student sees a clean equation written with an equals sign and answers yes.
\[ \lim_{x \to 0} \frac{1}{x^2} = \infty \]
Report that the limit exists and its value is the infinity symbol
Why: The notation looks exactly like every other limit statement, so the student treats the right-hand side as an answer and writes that the limit exists.
\[ \text{``the limit exists and equals } \infty \text{''} \ \times \]
The error, named
Why: The definition of a limit requires the outputs to get arbitrarily close to a real number L. The symbol on the right is not a real number, so no L has been produced. Treating it as one later leads to nonsense like subtracting it from itself and getting zero.
Read the statement as a description of behavior, and answer the existence question separately.
\[ \lim_{x \to 0} \frac{1}{x^2} = \infty \]
Say both halves out loud
Why: The limit does not exist, because no real number is being approached. The notation additionally tells you how it fails: the outputs grow without bound on both sides.
\[ \text{DNE, and specifically } \ \frac{1}{x^2} \to \infty \ \checkmark \]
Use it for what it is good for
Why: The infinite-limit statement is the sentence that establishes a vertical asymptote, and it tells you which way each branch of the graph runs. That is real information, and it is more than a bare statement that the limit does not exist.
| statement | what it buys you |
|---|---|
| the limit does not exist | no single number is approached |
| the outputs grow without bound | there is a vertical asymptote there |
| one side up, one side down | the two branches point opposite ways |
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Ranking
Put in order
These are the steps of Pattern: test a suspicious input, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
For a quotient, whenever the denominator is zero at the input you care about, run this decision procedure.
Step four is the one that separates a B from an A. Signs are where careful students still lose points.
Edge cases
Discussion prompt
Pattern: test a suspicious input works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
For a quotient, whenever the denominator is zero at the input you care about, run this decision procedure.
Prediction
Predict first
Which vertical line is a vertical asymptote of this function?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: the line where x equals negative 3
Why: The denominator factors as the quantity x minus 4 times the quantity x plus 3. The factor of x minus 4 cancels with the numerator, leaving 1 divided by the quantity x plus 3, so the input 4 is only a hole. The surviving factor blows up at negative 3, which is the vertical asymptote.
Check
Factor before you answer. The denominator has two zeros and they do not behave the same way.
\[ f(x) = \frac{x-4}{x^2 - x - 12} \]
Check your understanding
Which vertical line is a vertical asymptote of this function?
Answer: A
Why: The denominator factors as the quantity x minus 4 times the quantity x plus 3. The factor of x minus 4 cancels with the numerator, leaving 1 divided by the quantity x plus 3, so the input 4 is only a hole. The surviving factor blows up at negative 3, which is the vertical asymptote.
Section
Section 7
Concept
Every sentence in this deck has leaned on two vague phrases: arbitrarily close and close enough. Mathematics eventually has to cash those cheques.
\[ \lim_{x \to a} f(x) = L \quad \text{means:} \]
\[ \text{for every } \varepsilon > 0 \ \text{ there exists } \ \delta > 0 \ \text{ such that } \ 0 < |x - a| < \delta \ \Longrightarrow \ |f(x) - L| < \varepsilon \]
epsilon — The output tolerance. It is how close to the limit someone demands the outputs be, and it can be as small as they like.
delta — The input tolerance you supply in response. It is how close to the target the inputs must be in order to keep that promise.
Explain it
Discussion prompt
Explain The epsilon-delta definition, stated once to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Every sentence in this deck has leaned on two vague phrases: arbitrarily close and close enough. Mathematics eventually has to cash those cheques.
Intuition
Think of it as a two-player game. A skeptic goes first and names an output tolerance: get within one thousandth of the limit.
You answer with an input tolerance: stay within this distance of the target and I guarantee it. If you can always answer, no matter how brutal the demand, the limit is that number.
That is exactly what arbitrarily close meant all along. The skeptic moves first on purpose, because your answer is allowed to depend on their demand.
Notice the strict inequality on the left of the input condition. It is what excludes the target input itself, and it is why holes never matter.
Analogy
Discussion prompt
Explain The tolerance game by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Think of it as a two-player game. A skeptic goes first and names an output tolerance: get within one thousandth of the limit.
Picture it
Animation
Shows: The tolerance game — a rendered Manim animation.
Rendered with Manim.
Takeaway: If I can always answer, the limit really is L.
Step zero
Discussion prompt
Answering one tolerance demand — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write down exactly what the demand says about the outputs
Answer:
Worked example
For this line, the limit at the input two is clearly seven. A skeptic demands that the outputs be within one hundredth of seven. Find an input tolerance that delivers.
\[ f(x) = 3x + 1, \qquad a = 2, \qquad L = 7, \qquad \varepsilon = 0.01 \]
Write down exactly what the demand says about the outputs
Why: The demand is a distance statement. Absolute value is the right tool because it means distance regardless of direction.
\[ |f(x) - L| < 0.01 \ \Longleftrightarrow \ |(3x+1) - 7| < 0.01 \]
Simplify the inside and pull out the coefficient
Why: The expression inside collapses to three times the distance from the input to two, which is precisely the quantity delta is supposed to control.
\[ |3x - 6| = 3\,|x - 2| < 0.01 \]
Solve for the input distance to read off delta
Why: Dividing the demand by three converts an output tolerance into an input tolerance. Every step here is reversible, so this input condition genuinely forces the output condition.
\[ |x - 2| < \frac{0.01}{3} \quad \Rightarrow \quad \delta = \frac{1}{300} \approx 0.00333 \]
Note the general answer for any demand
Why: Nothing about the number one hundredth was special. The same algebra turns any tolerance into one third of itself, which is why this limit exists rather than merely looking plausible.
\[ \delta = \frac{\varepsilon}{3} \quad \text{works for every } \varepsilon > 0 \]
Verify with a specific input inside the tolerance
Why: Take the input 2.003, whose distance from two is 0.003, safely under one three-hundredth. Its output is 7.009, whose distance from seven is 0.009, safely under one hundredth. The promise holds.
\[ f(2.003) = 3(2.003) + 1 = 7.009, \quad |7.009 - 7| = 0.009 < 0.01 \ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Answering one tolerance demand", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Take the input 2.003, whose distance from two is 0.003, safely under one three-hundredth. Its output is 7.009, whose distance from seven is 0.009, safely under one hundredth. The promise holds.
Concept
You will not be asked to grind these proofs all semester. So why introduce it at all?
approaches a testable claim rather than a feeling, so limits can be proved instead of guessed.For the rest of Calculus I you will use the informal picture and the algebra. Just know that something rigorous is holding it up.
Counterexample
Discussion prompt
You will not be asked to grind these proofs all semester. So why introduce it at all?
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
For the rest of Calculus I you will use the informal picture and the algebra. Just know that something rigorous is holding it up.
Section
Section 8
Constraint
Discussion prompt
Run Pattern: the complete limit routine with this step confiscated:
Denominator zero and numerator not zero? Vertical asymptote. Do the sign analysis and report each side.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
Faced with any limit at a finite input, work through these in order and stop at the first one that answers the question.
And whatever the answer is, keep it separate from the function value. Those are two questions, and a good write-up answers both.
Real world
Discussion prompt
Outside this lesson: where does Limits: The Graphical and Numerical Idea actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: the complete limit routine is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
Two problems force calculus into existence: finding the slope of a tangent line, and finding instantaneous velocity. This deck starts there, then gives the informal definition of a limit and shows how to estimate limits from tables and read them off graphs. It covers one-sided limits and the two-sided existence test, removable holes, the three ways a limit can fail to exist, and infinite limits and vertical asymptotes, before closing with an honest first look at epsilon-delta. It targets the classic traps: assuming the limit is the function value, declaring that a limit exists when the one-sided limits disagree, calling an infinite limit an existing limit, and trusting a table on an oscillating function.
Ranking
Put in order
Put the moves of A full analysis at a seam into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Only inputs below three matter here, and for all of them the factor of x minus three is nonzero, so cancelling it is legal on the entire region the left-hand limit inspects.
Worked example
Find the two one-sided limits, the two-sided limit, and the function value at the input three.
\[ h(x) = \begin{cases} \dfrac{x^2 - 9}{x - 3}, & x < 3 \\[6pt] 4, & x = 3 \\[4pt] 2x, & x > 3 \end{cases} \]
Handle the left side by factoring the quotient piece
Why: Only inputs below three matter here, and for all of them the factor of x minus three is nonzero, so cancelling it is legal on the entire region the left-hand limit inspects.
\[ \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3 \ \Rightarrow \ \lim_{x \to 3^{-}} h(x) = 6 \]
Handle the right side with its own formula
Why: Inputs above three are governed by the doubling formula, which is a line with no trouble anywhere, so substitution is legal for that one-sided limit.
\[ \lim_{x \to 3^{+}} h(x) = \lim_{x \to 3^{+}} 2x = 6 \]
Apply the existence test
Why: The two one-sided limits both exist and both equal six, so by the existence test the two-sided limit exists and is six.
\[ \lim_{x \to 3} h(x) = 6 \]
Read the function value, which the middle line hands you directly
Why: The piecewise definition assigns the output four at the input three. The limit is six, so the two disagree and this point is a removable discontinuity.
\[ h(3) = 4 \neq 6 = \lim_{x \to 3} h(x) \]
Verify both approaches numerically
Why: At the input 2.999 the left formula gives 2.999 plus 3, which is 5.999. At the input 3.001 the right formula gives twice 3.001, which is 6.002. Both are within three thousandths of six, confirming the algebra, and neither is anywhere near the value four sitting at the point itself.
| x | formula in force | h(x) |
|---|---|---|
| 2.99 | x plus 3 (after cancelling) | 5.99 |
| 2.999 | x plus 3 (after cancelling) | 5.999 |
| 3 | the assigned value | 4 |
| 3.001 | 2x | 6.002 |
| 3.01 | 2x | 6.02 |
Picture it
Animation
Shows: Each line of the worked example "A full analysis at a seam", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the input 2.999 the left formula gives 2.999 plus 3, which is 5.999. At the input 3.001 the right formula gives twice 3.001, which is 6.002. Both are within three thousandths of six, confirming the algebra, and neither is anywhere near the value four sitting at the point itself.
Elimination
Eliminate the wrong options
What is the limit of f as x approaches 3?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The numerator factors as the quantity x minus 2 times the quantity x minus 3, so for every input other than 3 the function equals x minus 2. Letting the input close in on 3 from either side gives 3 minus 2, which is 1. The assigned value 7 sits only at the excluded point.
Check
Someone has deliberately assigned a value at the input three that does not match the surrounding behavior. Find the limit.
\[ f(x) = \begin{cases} \dfrac{x^2 - 5x + 6}{x - 3}, & x \neq 3 \\[6pt] 7, & x = 3 \end{cases} \]
Check your understanding
What is the limit of f as x approaches 3?
Answer: A
Why: The numerator factors as the quantity x minus 2 times the quantity x minus 3, so for every input other than 3 the function equals x minus 2. Letting the input close in on 3 from either side gives 3 minus 2, which is 1. The assigned value 7 sits only at the excluded point.
Check
A student tabulates the function below at the inputs 1, then 0.1, then 0.01, then 0.001, and gets an output of zero every single time. They conclude the limit is zero.
\[ f(x) = \sin\!\left(\frac{\pi}{x}\right), \qquad x \to 0 \]
Check your understanding
What is actually true about this limit?
Answer: A
Why: Every input the student chose made the angle a whole multiple of half a turn, where the sine is exactly zero. But inputs like 0.4 and two divided by nine make the angle an odd multiple of a quarter turn, where the sine is 1 or negative 1, and such inputs exist arbitrarily close to zero. The outputs never settle.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Why Calculus Needs Limits · What a Limit Actually Says · Estimating a Limit from a Table · Reading Limits Off a Graph · When a Limit Fails to Exist · Infinite Limits and Vertical Asymptotes. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A limit answers one question and one question only: where are the outputs headed as the inputs close in? Not where they land, and not what happens at the point.
| what you see | what to do | what to report |
|---|---|---|
| substitution gives a number | you are done | that number |
| zero over zero | factor, cancel, or rationalize | the simplified limit |
| nonzero over zero | sign analysis on each side | asymptote, one side at a time |
| a seam or an absolute value | both one-sided limits | existence test verdict |
| endless wiggling | two families of inputs | does not exist |
Next up: the limit laws, which let you break a complicated limit into simple pieces, plus the algebraic techniques that turn the zero-over-zero cases into one-line answers.
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