This deck covers the limit laws and the conditions attached to each of them, then works out when direct substitution is legal and what to do when it is not. It handles the zero-over-zero indeterminate form with factoring and cancelling, conjugates, and complex fractions, takes piecewise limits at the seam, and finishes with the Squeeze Theorem and the classic sine-over-x limit. It targets the traps of cancelling a factor and forgetting the hole, treating zero over zero as automatically zero or one, using the quotient law when the bottom limit is zero, and multiplying by a conjugate on only part of the fraction.
Subject: Calculus I · 135 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 02
Computing limits exactly: substitution, factoring, conjugates, piecewise seams, and the Squeeze Theorem
Objectives
Deck 1 taught you to see a limit on a graph or in a table. This deck teaches you to compute one exactly.
Warm-up
Discussion prompt
Before we open Limit Laws and Computing Limits Algebraically: without looking back, what was the main idea of Limits: The Graphical and Numerical Idea, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
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.
Section
Part 1
Concept
Estimating from a table gets you close. Squinting at a graph gets you close. Neither gives you an exact answer, and neither scales to a function nobody has graphed for you.
The limit laws are the bridge: they let you break a complicated limit into pieces you already know, compute each piece, and reassemble.
Counterexample
Discussion prompt
Estimating from a table gets you close. Squinting at a graph gets you close. Neither gives you an exact answer, and neither scales to a function nobody has graphed for you.
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:
The limit laws are the bridge: they let you break a complicated limit into pieces you already know, compute each piece, and reassemble.
Intuition
Two runners are approaching the same finish line. One is settling toward a pace of 3, the other toward a pace of 5. Where is their combined pace settling? Toward 8. You did not need the exact formulas.
That is the whole idea. If two pieces are each settling somewhere, then their sum, difference, product, and quotient settle at exactly the place you would predict.
The limit laws are that sentence, written carefully, with the fine print about when it can fail.
Analogy
Discussion prompt
Explain Limits respect arithmetic 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:
The limit laws are that sentence, written carefully, with the fine print about when it can fail.
Concept
Two limits are so basic they need no proof machinery. Every other limit law is assembled from these.
\[ \lim_{x \to a} c = c \]
A constant function never moves, so it is already sitting at its limit.
\[ \lim_{x \to a} x = a \]
As the input creeps toward a number, the output of the identity function is the input, so it creeps toward the same number.
Explain it
Discussion prompt
Explain The two limits everything is built from 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:
Two limits are so basic they need no proof machinery. Every other limit law is assembled from these.
Picture it
Animation
Shows: The two limits everything is built from — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every other law is assembled out of these two.
Concept
Suppose both of these limits exist as real numbers.
\[ \lim_{x \to a} f(x) = L \qquad \lim_{x \to a} g(x) = M \]
\[ \lim_{x \to a} \left[ f(x) + g(x) \right] = L + M \]
\[ \lim_{x \to a} \left[ f(x) - g(x) \right] = L - M \]
In words: the limit of a sum is the sum of the limits. You may split a sum apart, handle the pieces, and add the answers.
Concept
A constant factor is just along for the ride. Pull it outside the limit.
\[ \lim_{x \to a} \left[ c \cdot f(x) \right] = c \cdot L \]
And a product of two settling quantities settles at the product.
\[ \lim_{x \to a} \left[ f(x) \, g(x) \right] = L \cdot M \]
Concept
Division is the only arithmetic operation that can fail, so this law is the only one that carries a condition.
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M} \quad \text{provided } M \neq 0 \]
the quotient condition — The denominator's limit must be a nonzero number. If the bottom limit is zero, this law says nothing at all - it does not say the answer is zero, and it does not say the limit fails to exist. It simply does not apply.
Circle that condition. Almost every interesting limit in this deck is a case where it is violated, and the rest of the deck is about what to do then.
Picture it
Animation
Shows: The quotient law, and its one condition — a rendered Manim animation.
Rendered with Manim.
Takeaway: That condition is exactly where the interesting problems live.
Concept
Applying the product law to a function times itself, over and over, gives the power law.
\[ \lim_{x \to a} \left[ f(x) \right]^{n} = L^{n} \]
Roots work the same way, with the obvious restriction that you cannot take an even root of a negative number.
\[ \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L} \quad (\text{if } L > 0 \text{ when } n \text{ is even}) \]
Picture it
Animation
Shows: Powers and roots pass through — a rendered Manim animation.
Rendered with Manim.
Takeaway: The root law needs the inside to be legal at the limit.
Concept
Every law above begins with the same clause: assume both individual limits exist as real numbers. That clause is not decoration.
The laws run in one direction only. If the two pieces have limits, the combination does. The reverse is not guaranteed: a sum can settle down even when neither piece does.
\[ \lim_{x \to 0}\left[\sin\!\left(\tfrac{1}{x}\right) - \sin\!\left(\tfrac{1}{x}\right)\right] = \lim_{x \to 0} 0 = 0 \]
Each piece there oscillates forever and has no limit, yet the difference is the constant zero. So you may not run the sum law backwards to claim each piece has a limit.
Picture it
Animation
Shows: The sum, product and quotient limit laws written out.
The condition is the same every time.
Takeaway: Every law lets you break a limit into pieces, and every one carries the same fine print: the pieces must themselves exist before you may combine them.
Ranking
Put in order
Put the moves of Worked example: a polynomial, law by law 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. Three terms are being added and subtracted, and each term will turn out to have a limit, so the law applies.
Worked example
Evaluate this using only the laws, one step at a time.
\[ \lim_{x \to 3} \left( 2x^{2} - 5x + 1 \right) \]
Split the expression across the sum and difference laws
Why: Three terms are being added and subtracted, and each term will turn out to have a limit, so the law applies.
\[ = \lim_{x \to 3} 2x^{2} - \lim_{x \to 3} 5x + \lim_{x \to 3} 1 \]
Pull the constants 2 and 5 outside their limits
Why: The constant multiple law: a fixed factor does not change where the rest is heading.
\[ = 2\lim_{x \to 3} x^{2} - 5\lim_{x \to 3} x + \lim_{x \to 3} 1 \]
Use the power law on the squared term and the two building-block limits on the rest
Why: The limit of the identity function is 3, so its square has limit 3 squared; the limit of the constant 1 is 1.
\[ = 2(3)^{2} - 5(3) + 1 = 18 - 15 + 1 = 4 \]
Verify by evaluating the original polynomial near 3
Why: At 2.99 the value is 2(8.9401) - 14.95 + 1 = 3.9302, and at 3.01 it is 2(9.0601) - 15.05 + 1 = 4.0702. The outputs close in on 4 from both sides.
| x | value of the polynomial |
|---|---|
| 2.9 | 3.6200 |
| 2.99 | 3.9302 |
| 3.01 | 4.0702 |
| 3.1 | 4.7200 |
Picture it
Animation
Shows: A polynomial, law by law — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every polynomial limit collapses to substitution, and the laws are why.
Step zero
Discussion prompt
Worked example: a limit under a root — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the limit of the inside expression first
Answer:
Worked example
The root law lets you work under the radical first.
\[ \lim_{x \to 5} \sqrt{2x^{2} - 14} \]
Find the limit of the inside expression first
Why: The root law says the limit passes inside the radical, provided the inside limit is not negative under an even root.
\[ \lim_{x \to 5}\left(2x^{2} - 14\right) = 2(25) - 14 = 36 \]
Take the square root of that inside limit
Why: The inside limit 36 is positive, so the even-root restriction is satisfied and the law applies.
\[ \lim_{x \to 5} \sqrt{2x^{2} - 14} = \sqrt{36} = 6 \]
Verify with values on both sides of 5
Why: At 4.99 the inside is 2(24.9001) - 14 = 35.8002 and its root is about 5.98333; at 5.01 the inside is 36.2002 and its root is about 6.01666. Both sit right beside 6.
| x | inside expression | square root |
|---|---|---|
| 4.99 | 35.8002 | 5.98333 |
| 5.01 | 36.2002 | 6.01666 |
Concept
Chaining the laws for every polynomial would be exhausting. The good news is that the laws prove a shortcut once and for all.
direct substitution property — If f is a polynomial, or a rational function whose denominator is not zero at a, then the limit of f as x approaches a is exactly f(a). You may simply plug the number in.
The same shortcut works for roots, sines, cosines, exponentials, and logarithms, as long as the number you are approaching is genuinely inside the domain.
Picture it
Animation
Shows: A polynomial limit evaluated by direct substitution.
Start here every single time.
Takeaway: For a continuous function, substitution simply is the answer. Most limits are this easy, and only the ones that break need technique.
Intuition
A limit asks where the outputs are heading. The function value asks where the output is. Those are different questions, and Deck 1 showed they can disagree.
They agree exactly when the graph has no break, no hole, and no jump at that spot - when you could draw it through the point without lifting your pencil.
Polynomials are the smoothest curves there are. Nothing is ever missing. So heading-toward and sitting-at are the same number, and substitution is legal. Deck 3 gives this idea its real name: continuity.
Estimation
Predict first
Always try substitution first. Here it succeeds on the first try.
Commit before you compute: what does Worked example: a rational function where substitution works 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 numerically from both sides of 2
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 1.99 the quotient is 8.9301 divided by 5.99, about 1.490835; at 2.01 it is 9.0701 divided by 6.01, about 1.509168.
Worked example
Always try substitution first. Here it succeeds on the first try.
\[ \lim_{x \to 2} \frac{x^{2} + 3x - 1}{x + 4} \]
Check the denominator at the target number before anything else
Why: The quotient law needs a nonzero bottom limit. At 2 the denominator is 2 plus 4, which is 6, so the condition holds and substitution is legal.
Substitute 2 into the numerator and the denominator
Why: Both top and bottom are polynomials, so each one's limit is its value at 2.
\[ = \frac{(2)^{2} + 3(2) - 1}{2 + 4} = \frac{4 + 6 - 1}{6} = \frac{9}{6} \]
Reduce the fraction to lowest terms
Why: Both 9 and 6 share a factor of 3; a test answer should be fully simplified.
\[ \lim_{x \to 2} \frac{x^{2} + 3x - 1}{x + 4} = \frac{3}{2} \]
Verify numerically from both sides of 2
Why: At 1.99 the quotient is 8.9301 divided by 5.99, about 1.490835; at 2.01 it is 9.0701 divided by 6.01, about 1.509168. Both squeeze in on 1.5, which is three halves.
| x | numerator | denominator | quotient |
|---|---|---|---|
| 1.99 | 8.9301 | 5.99 | 1.490835 |
| 2.01 | 9.0701 | 6.01 | 1.509168 |
Picture it
Animation
Shows: Each line of the worked example "a rational function where substitution works", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The quotient law needs a nonzero bottom limit. At 2 the denominator is 2 plus 4, which is 6, so the condition holds and substitution is legal.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The tempting move: apply the quotient law anyway and let the zero take care of itself.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Backwards. Dividing by a tiny number makes the result enormous, not tiny.
First check the condition the law actually carries.
Why: Backwards. Dividing by a tiny number makes the result enormous, not tiny. Dividing 3 by one hundredth gives 300.
Trap
The tempting move: apply the quotient law anyway and let the zero take care of itself.
\[ \lim_{x \to 1} \frac{x + 2}{x - 1} \;\overset{?}{=}\; \frac{3}{0} = 0 \]
The reasoning is: a small denominator makes a small fraction, so the answer is zero
Why: Backwards. Dividing by a tiny number makes the result enormous, not tiny. Dividing 3 by one hundredth gives 300.
The real values do the opposite of settling to zero
Why: Just to the right of 1 the quotient rockets upward; just to the left it plunges. Nothing is approaching zero.
| x | value of the quotient |
|---|---|
| 0.99 | -299 |
| 0.999 | -2999 |
| 1.001 | 3001 |
| 1.01 | 301 |
First check the condition the law actually carries.
\[ \lim_{x \to 1}(x - 1) = 0 \;\Rightarrow\; \text{the quotient law does not apply} \]
Look at each side separately instead
Why: With a nonzero top and a bottom shrinking to zero, the sign of the bottom decides everything, and that sign is different on the two sides.
\[ \lim_{x \to 1^{-}} \frac{x+2}{x-1} = -\infty, \qquad \lim_{x \to 1^{+}} \frac{x+2}{x-1} = +\infty \]
State the conclusion honestly
Why: The two-sided limit does not exist, and the line at 1 is a vertical asymptote. That is a description of the failure, not a numeric answer.
Pattern
Step through it
Step through Trap: using the quotient law when the bottom limit is zero one row at a time. What is driving the change, and what would the row after the last one be?
Pattern
Every limit problem in this course starts the same way. Plug the number in and look at what comes out. There are only three possible outcomes.
| What substitution gives | What it means | What to do |
|---|---|---|
| A real number | Substitution was legal | You are done. That is the limit. |
| A nonzero number over zero | Denominator is collapsing, top is not | Check each side; expect an infinite limit or a two-sided failure. |
| Zero over zero | Indeterminate - no information yet | Do algebra: factor, conjugate, or clear fractions. Then substitute again. |
The rest of this deck is the third row: the toolbox for the zero-over-zero case.
Comparison
Comparison matrix
From Pattern: substitute first, then read the result: refill the What it means column from what you know. The rest of the table is as it appeared.
| What substitution gives | What it means | What to do |
|---|---|---|
| A real number | Substitution was legal | You are done. That is the limit. |
| A nonzero number over zero | Denominator is collapsing, top is not | Check each side; expect an infinite limit or a two-sided failure. |
| Zero over zero | Indeterminate - no information yet | Do algebra: factor, conjugate, or clear fractions. Then substitute again. |
Check
Evaluate the denominator first, then decide. Work it before you click.
\[ \lim_{x \to -2} \frac{x^{3} + 5x}{x^{2} - 1} \]
Check your understanding
What is the value of this limit?
Answer: A
Why: The denominator at -2 is 4 minus 1, which is 3 and not zero, so substitution is legal. The numerator is -8 plus -10, which is -18, and -18 divided by 3 is -6.
Section
Part 2
Concept
When substitution produces this, many students want to write down an answer immediately. There is nothing to write down yet.
\[ \frac{0}{0} \]
indeterminate form — A form such as zero over zero that carries no information about the limit. Different functions producing this same form have different limits, so the form by itself can never decide the answer.
Read it as a message from the algebra: the top and the bottom are both being dragged to zero by a shared cause. Find the shared cause, remove it, and try again.
Picture it
Animation
Shows: A rational limit resolved by factoring and cancelling.
An instruction, not a result.
Takeaway: The indeterminate form is not an answer — it is an instruction to do algebra. The shared factor causing the trouble is always there to be cancelled.
Intuition
Here are three limits. Substituting zero into all three gives exactly the same form, zero over zero. Watch what actually happens.
\[ \lim_{x \to 0} \frac{3x}{x} = \lim_{x \to 0} 3 = 3 \]
\[ \lim_{x \to 0} \frac{x^{2}}{x} = \lim_{x \to 0} x = 0 \]
\[ \lim_{x \to 0} \frac{x}{x^{2}} = \lim_{x \to 0} \frac{1}{x} \quad \text{does not exist} \]
Same form, three different fates. The answer depends on which zero is winning the race, and only the algebra can tell you that.
| Expression | Form at zero | Actual limit |
|---|---|---|
| three x over x | zero over zero | 3 |
| x squared over x | zero over zero | 0 |
| x over x squared | zero over zero | does not exist |
Trade off
Comparison matrix
From Why the form cannot decide the answer: every row here is a choice with a cost. Fill the Form at zero column, then say which row you would actually pick and what you give up for it.
| Expression | Form at zero | Actual limit |
|---|---|---|
| three x over x | zero over zero | 3 |
| x squared over x | zero over zero | 0 |
| x over x squared | zero over zero | does not exist |
Anomaly
Predict first
A student writes this, and it looks reasonable:
The two most common guesses: anything over itself is one, or zero on top makes the whole thing zero.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Zero over zero is not a division you can perform.
Treat the form as an instruction to do algebra. Factor the shared cause out of the top.
Why: Zero over zero is not a division you can perform. Nothing is actually being divided until you simplify the expression.
Trap
The two most common guesses: anything over itself is one, or zero on top makes the whole thing zero.
\[ \lim_{x \to 0} \frac{x^{2} + 5x}{x} \;\overset{?}{=}\; \frac{0}{0} = 0 \quad \text{or} \quad 1 \]
Both guesses come from applying an arithmetic habit to an expression that is not arithmetic
Why: Zero over zero is not a division you can perform. Nothing is actually being divided until you simplify the expression.
The real values are near neither guess
Why: The outputs are closing in on 5, which is neither of the two guesses. Verified by direct evaluation on both sides.
| x | value of the quotient |
|---|---|
| -0.01 | 4.99 |
| -0.001 | 4.999 |
| 0.001 | 5.001 |
| 0.01 | 5.01 |
Treat the form as an instruction to do algebra. Factor the shared cause out of the top.
\[ \frac{x^{2} + 5x}{x} = \frac{x(x + 5)}{x} \]
Cancel the shared factor of x
Why: In a limit as x approaches 0, x is never actually 0, so dividing by it is legal at every point we care about.
\[ \lim_{x \to 0} \frac{x^{2} + 5x}{x} = \lim_{x \to 0} (x + 5) = 5 \]
Notice the answer matches the table
Why: The table closed in on 5 from both sides, exactly as the algebra predicts. The form gave no information; the factoring gave all of it.
Pattern
Step through it
Step through Trap: treating zero over zero as automatically zero or one one row at a time. What is driving the change, and what would the row after the last one be?
Concept
For a rational function, the zero-over-zero form is not bad luck. It is a guarantee that a common factor exists.
If substituting the target number makes the top zero, then that number is a root of the top polynomial. Same for the bottom.
factor theorem — If a polynomial p has p(a) equal to zero, then x minus a divides p exactly. A root and a linear factor are two views of the same fact.
\[ p(a) = 0 \text{ and } q(a) = 0 \;\Rightarrow\; (x - a) \text{ divides both } p \text{ and } q \]
So the shared cause is the factor for the number you are approaching. Cancel it, and the collapse disappears.
Picture it
Animation
Shows: Why a shared factor is always there — a rendered Manim animation.
Rendered with Manim.
Takeaway: The Factor Theorem guarantees the cancellation exists before you look for it.
Pattern
Predict first
The table runs: 2.9 | 5.9 · 2.99 | 5.99 · 3.01 | 6.01
In Worked example: factor and cancel, given the rows so far: what is the next one — the row where x is 3.1?
Correct: 3.1 | 6.1
| x | original quotient |
|---|---|
| 2.9 | 5.9 |
| 2.99 | 5.99 |
| 3.01 | 6.01 |
| 3.1 | 6.1 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. The numerator becomes 9 minus 9, and the denominator becomes 3 minus 3.
Worked example
The starting move never changes: substitute and look.
\[ \lim_{x \to 3} \frac{x^{2} - 9}{x - 3} \]
Substitute 3 and read the result
Why: The numerator becomes 9 minus 9, and the denominator becomes 3 minus 3. Both are zero, so this is the indeterminate form and more work is required.
Factor the numerator as a difference of squares
Why: Since 3 is a root of the top, the factor theorem promises a factor of x minus 3, and the difference-of-squares pattern hands it to you.
\[ \frac{x^{2} - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} \]
Cancel the factor of x minus 3
Why: The limit never evaluates the function at 3 itself, so x minus 3 is a nonzero number throughout, and cancelling it is legal.
\[ \lim_{x \to 3} \frac{x^{2}-9}{x-3} = \lim_{x \to 3} (x + 3) = 6 \]
Verify with the original quotient on both sides of 3
Why: At 2.99 the quotient is -0.0599 divided by -0.01, which is 5.99; at 3.01 it is 0.0601 divided by 0.01, which is 6.01. The original expression, not the simplified one, closes in on 6.
| x | original quotient |
|---|---|
| 2.9 | 5.9 |
| 2.99 | 5.99 |
| 3.01 | 6.01 |
| 3.1 | 6.1 |
Picture it
Animation
Shows: Each line of the worked example "factor and cancel", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 2.99 the quotient is -0.0599 divided by -0.01, which is 5.99; at 3.01 it is 0.0601 divided by 0.01, which is 6.01. The original expression, not the simplified one, closes in on 6.
Concept
Cancelling did not simplify the function. It replaced it with a different function that happens to agree everywhere except at one point.
\[ \frac{x^{2}-9}{x-3} = x + 3 \quad \text{for every } x \neq 3 \]
At 3 the left side is undefined and the right side equals 6. Everywhere else they are identical.
That is precisely why the swap is legal inside a limit: a limit only inspects the nearby points, never the point itself. Two functions that agree near a point have the same limit there.
Trap
After cancelling, it is tempting to declare that the function simply is a line now.
\[ f(x) = \frac{x^{2}-9}{x-3} = x + 3 \;\Rightarrow\; f(3) = 6 \;? \]
The claim slides from a statement about the limit to a statement about the value
Why: Those are different questions. Cancelling was only ever justified for inputs that are not 3.
Ask what the original formula actually returns at 3
Why: It returns zero divided by zero, which is not a number. The function has no output at 3 at all.
| Question | Answer |
|---|---|
| limit as x approaches 3 | 6 |
| value of f at 3 | undefined |
| domain of f | all real numbers except 3 |
Keep the two claims separate and attach the condition to the cancellation.
\[ f(x) = x + 3 \quad \text{for } x \neq 3, \qquad \lim_{x \to 3} f(x) = 6 \]
Describe the graph in one sentence
Why: It is the straight line through the usual points, with a single point punched out at height 6 above the input 3.
Say what the limit is doing there
Why: The limit is 6 because the nearby outputs surround 6; the hole is invisible to the limit. This is the removable discontinuity from Deck 1, seen from the algebra side.
Comparison
Comparison matrix
From Trap: cancelling the factor and forgetting the hole: refill the Answer column from what you know. The rest of the table is as it appeared.
| Question | Answer |
|---|---|
| limit as x approaches 3 | 6 |
| value of f at 3 | undefined |
| domain of f | all real numbers except 3 |
Fill the middle
Fill in the blanks
From Worked example: factoring both the top and the bottom — finish the line. Write what belongs on the right of the equals sign before you look.
x^(x+2)(x-1) + x - 2 = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The top is 4 minus 2 minus 2, which is zero; the bottom is 4 minus 4, which is zero.
Worked example
Now both polynomials need factoring.
\[ \lim_{x \to -2} \frac{x^{2} + x - 2}{x^{2} - 4} \]
Substitute negative 2 first
Why: The top is 4 minus 2 minus 2, which is zero; the bottom is 4 minus 4, which is zero. Indeterminate, so factor.
Factor the numerator into two binomials
Why: Two numbers multiplying to negative 2 and adding to 1 are 2 and negative 1, and the expanded product checks out.
\[ x^{2} + x - 2 = (x+2)(x-1) \]
Factor the denominator as a difference of squares
Why: Both factorizations contain x plus 2, which is exactly the factor the factor theorem promised for the root negative 2.
\[ \frac{(x+2)(x-1)}{(x+2)(x-2)} = \frac{x-1}{x-2} \quad (x \neq -2) \]
Substitute negative 2 into the reduced expression
Why: The reduced denominator at negative 2 is negative 4, which is not zero, so substitution is now legal.
\[ \lim_{x \to -2} \frac{x^{2}+x-2}{x^{2}-4} = \frac{-3}{-4} = \frac{3}{4} \]
Verify with the original expression on both sides
Why: At negative 1.99 the original quotient is -0.0299 over -0.0399, about 0.749373; at negative 2.01 it is 0.0301 over 0.0401, about 0.750623. Both bracket 0.75, which is three quarters.
| x | original quotient |
|---|---|
| -1.99 | 0.749373 |
| -2.01 | 0.750623 |
Picture it
Animation
Shows: Each line of the worked example "factoring both the top and the bottom", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At negative 1.99 the original quotient is -0.0299 over -0.0399, about 0.749373; at negative 2.01 it is 0.0301 over 0.0401, about 0.750623. Both bracket 0.75, which is three quarters.
Step zero
Discussion prompt
Worked example: a cubic on top — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Substitute 1 and confirm the indeterminate form
Answer:
Worked example
The same plan works when the numerator is a cubic, as long as you know the pattern.
\[ \lim_{x \to 1} \frac{x^{3} - 1}{x^{2} - 1} \]
Substitute 1 and confirm the indeterminate form
Why: The top is 1 minus 1 and the bottom is 1 minus 1, so both collapse to zero and a shared factor of x minus 1 must exist.
Factor the top with the difference-of-cubes pattern
Why: A cube minus a cube factors as the difference of the bases times the sum of the squares and the cross term. Expanding confirms it returns the cubic.
\[ x^{3} - 1 = (x-1)\left(x^{2} + x + 1\right) \]
Factor the bottom and cancel
Why: The bottom is a difference of squares, so the shared factor x minus 1 appears in both and divides out.
\[ \frac{(x-1)(x^{2}+x+1)}{(x-1)(x+1)} = \frac{x^{2}+x+1}{x+1} \]
\[ \lim_{x \to 1} \frac{x^{3}-1}{x^{2}-1} = \frac{1+1+1}{1+1} = \frac{3}{2} \]
Verify numerically with the original quotient
Why: At 0.99 the quotient is -0.029701 over -0.0199, about 1.492513; at 1.01 it is 0.030301 over 0.0201, about 1.507512. Both close in on 1.5, which is three halves.
| x | top | bottom | quotient |
|---|---|---|---|
| 0.99 | -0.029701 | -0.0199 | 1.492513 |
| 1.01 | 0.030301 | 0.0201 | 1.507512 |
Picture it
Animation
Shows: Each line of the worked example "a cubic on top", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 0.99 the quotient is -0.029701 over -0.0199, about 1.492513; at 1.01 it is 0.030301 over 0.0201, about 1.507512. Both close in on 1.5, which is three halves.
Check
Substitute first, confirm the form, then factor both polynomials.
\[ \lim_{x \to 4} \frac{x^{2} - 16}{x^{2} - 3x - 4} \]
Check your understanding
What is the value of this limit?
Answer: A
Why: Both parts vanish at 4. Factoring gives the top as x minus 4 times x plus 4 and the bottom as x minus 4 times x plus 1. Cancelling leaves x plus 4 over x plus 1, which at 4 is 8 over 5.
Section
Part 3
Concept
Factoring only helps when the top and bottom are polynomials. Put a square root in the expression and the shared factor is hidden where factoring cannot reach it.
\[ \lim_{x \to 0} \frac{\sqrt{x+9} - 3}{x} \]
Substituting gives zero over zero again, so a shared cause is still in there. You just need a different tool to expose it.
Intuition
There is exactly one algebra identity that makes square roots disappear: a difference times a sum.
\[ (A - B)(A + B) = A^{2} - B^{2} \]
If A is a square root, then squaring it deletes the radical. So multiply the radical expression by its conjugate - the same two terms with the sign between them flipped.
\[ \left(\sqrt{x+9} - 3\right)\left(\sqrt{x+9} + 3\right) = (x+9) - 9 = x \]
The root is gone and a bare factor of the very thing going to zero has appeared. That factor is the shared cause, and now it can cancel.
Picture it
Animation
Shows: A limit with a square root resolved by multiplying by the conjugate.
Multiplying by one, deliberately.
Takeaway: Multiplying by the conjugate over itself is multiplying by one, chosen in exactly the shape that clears the root and exposes the cancelling factor.
Missing information
Discussion prompt
The standard setup: a root minus a number on top, and the variable alone on the bottom.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The top is the square root of 9 minus 3, which is zero, and the bottom is zero. Indeterminate, so algebra is required.
Worked example
The standard setup: a root minus a number on top, and the variable alone on the bottom.
\[ \lim_{x \to 0} \frac{\sqrt{x+9} - 3}{x} \]
Substitute 0 and name the form
Why: The top is the square root of 9 minus 3, which is zero, and the bottom is zero. Indeterminate, so algebra is required.
Multiply by the conjugate over itself, top and bottom
Why: That expression equals 1, so the value of the fraction is untouched. Multiplying only one part would change the function into a different one.
\[ \frac{\sqrt{x+9}-3}{x} \cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} \]
Expand only the numerator and leave the denominator in factored form
Why: The whole point is to cancel, so multiplying out the bottom would hide the factor you are about to use.
\[ = \frac{(x+9) - 9}{x\left(\sqrt{x+9}+3\right)} = \frac{x}{x\left(\sqrt{x+9}+3\right)} \]
Cancel the shared factor of x
Why: In the limit as x approaches 0, x is never actually 0, so this division is legal at every nearby point.
\[ \lim_{x \to 0} \frac{1}{\sqrt{x+9}+3} = \frac{1}{3+3} = \frac{1}{6} \]
Verify with the original expression near zero
Why: At 0.01 the top is about 0.0016662 and dividing by 0.01 gives about 0.166621; at negative 0.01 the quotient is about 0.166713. Both bracket one sixth, about 0.166667.
| x | original quotient |
|---|---|
| -0.01 | 0.166713 |
| 0.01 | 0.166621 |
Picture it
Animation
Shows: Each line of the worked example "conjugate on the numerator", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 0.01 the top is about 0.0016662 and dividing by 0.01 gives about 0.166621; at negative 0.01 the quotient is about 0.166713. Both bracket one sixth, about 0.166667.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The shortcut that feels harmless: multiply the top by the conjugate and leave the bottom alone.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The conjugate near zero is close to 6, so the top was scaled by about 6 while the bottom was not.
Multiply by a genuine form of the number 1: the conjugate over itself.
Why: The conjugate near zero is close to 6, so the top was scaled by about 6 while the bottom was not. That is a different function, not a rewrite.
Trap
The shortcut that feels harmless: multiply the top by the conjugate and leave the bottom alone.
\[ \frac{\sqrt{x+9}-3}{x} \;\overset{?}{\to}\; \frac{(x+9)-9}{x} = \frac{x}{x} = 1 \]
This multiplied the function by roughly 6 without permission
Why: The conjugate near zero is close to 6, so the top was scaled by about 6 while the bottom was not. That is a different function, not a rewrite.
Compare the claim against real values
Why: The claimed answer is 1, but the original expression is nowhere near 1 at any nearby input. It is near 0.1667, which is one sixth.
| x | claimed value | true value |
|---|---|---|
| -0.01 | 1 | 0.166713 |
| 0.01 | 1 | 0.166621 |
Multiply by a genuine form of the number 1: the conjugate over itself.
\[ \frac{\sqrt{x+9}-3}{x} \cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} = \frac{x}{x\left(\sqrt{x+9}+3\right)} \]
Cancel and substitute
Why: The new denominator carries the conjugate factor, which is what makes the answer one sixth rather than 1.
\[ \lim_{x \to 0}\frac{1}{\sqrt{x+9}+3} = \frac{1}{6} \]
Sanity-check the size of the answer
Why: Six is the size of the conjugate at the target point, and one sixth is about 0.1667, exactly what the table of real values showed.
Notation
Annotate
From Trap: conjugating only part of the fraction — read this one piece at a time. What is each part doing?
On: \( \frac{\sqrt{x+9}-3}{x} \;\overset{?}{\to}\; \frac{(x+9)-9}{x} = \frac{x}{x} = 1 \)
Hypothesis
Predict first
Worked example: the root is on the bottom 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: Substitute 4 and confirm the indeterminate form
Why: The top is zero and the bottom is the square root of 4 minus 2, also 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
Nothing changes when the radical sits in the denominator. Conjugate the part that has the root.
\[ \lim_{x \to 4} \frac{x - 4}{\sqrt{x} - 2} \]
Substitute 4 and confirm the indeterminate form
Why: The top is zero and the bottom is the square root of 4 minus 2, also zero.
Multiply top and bottom by the conjugate of the denominator
Why: Flipping the sign gives the square root of x plus 2, and the product of the two is x minus 4, which clears the radical entirely.
\[ \frac{x-4}{\sqrt{x}-2} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{(x-4)\left(\sqrt{x}+2\right)}{x - 4} \]
Cancel the factor of x minus 4
Why: Near 4 but not at 4, this factor is a nonzero number, so it divides out cleanly and the expression becomes a simple sum.
\[ \lim_{x \to 4} \left(\sqrt{x} + 2\right) = 2 + 2 = 4 \]
Verify with the original quotient on both sides of 4
Why: At 3.99 the quotient is negative 0.01 over negative 0.00250156, about 3.99750; at 4.01 it is 0.01 over 0.00249844, about 4.00250. Both land beside 4.
| x | square root of x | original quotient |
|---|---|---|
| 3.99 | 1.99749844 | 3.99750 |
| 4.01 | 2.00249844 | 4.00250 |
Concept
A third shape shows up constantly, especially once you meet the definition of the derivative: a fraction whose numerator contains fractions.
complex fraction — A fraction that has one or more fractions inside its numerator or denominator. It is simplified by combining the inner fractions over a single common denominator, then multiplying by the reciprocal of the outer denominator.
Do not try to cancel anything until the numerator is one single fraction. The shared factor is invisible while the top is still a sum of pieces.
Picture it
Animation
Shows: Clear the small fractions first — a rendered Manim animation.
Rendered with Manim.
Takeaway: Multiply through by the common denominator and the mess collapses.
Ranking
Put in order
Put the moves of Worked example: a complex fraction 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 top is one half minus one half, which is zero, and the bottom is zero.
Worked example
This is the difference quotient of the reciprocal function, and it will reappear in Deck 5.
\[ \lim_{x \to 2} \frac{\dfrac{1}{x} - \dfrac{1}{2}}{x - 2} \]
Substitute 2 and confirm the form
Why: The top is one half minus one half, which is zero, and the bottom is zero. Indeterminate again.
Combine the two inner fractions over the common denominator two x
Why: One single fraction on top is the only way a cancellable factor can appear.
\[ \frac{1}{x} - \frac{1}{2} = \frac{2 - x}{2x} \]
Divide by x minus 2 by multiplying by its reciprocal
Why: Dividing by a quantity is multiplying by one over that quantity; this stacks everything into a single fraction.
\[ \frac{2-x}{2x} \cdot \frac{1}{x-2} = \frac{-(x-2)}{2x(x-2)} \]
Factor the negative sign out of two minus x, then cancel
Why: Two minus x is the opposite of x minus 2, so writing it that way exposes the shared factor and leaves a minus sign behind.
\[ \lim_{x \to 2} \frac{-1}{2x} = \frac{-1}{4} \]
Verify numerically on both sides of 2
Why: At 1.99 the top is 0.002512563 and dividing by negative 0.01 gives about negative 0.2512563; at 2.01 the result is about negative 0.2487562. Both bracket negative 0.25.
| x | top of the complex fraction | whole expression |
|---|---|---|
| 1.99 | 0.002512563 | -0.2512563 |
| 2.01 | -0.002487562 | -0.2487562 |
Picture it
Animation
Shows: Each line of the worked example "a complex fraction", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 1.99 the top is 0.002512563 and dividing by negative 0.01 gives about negative 0.2512563; at 2.01 the result is about negative 0.2487562. Both bracket negative 0.25.
Check
Confirm the form, multiply top and bottom by the conjugate, cancel, then substitute.
\[ \lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x} \]
Check your understanding
What is the value of this limit?
Answer: A
Why: Multiplying by the conjugate makes the numerator 4 plus x minus 4, which is x, so the expression becomes 1 over the square root of 4 plus x, plus 2. At 0 that is 1 over 4.
Section
Part 4
Concept
A piecewise function is one rule on one stretch of the number line and a different rule on another. The number where the rule changes is the seam.
Away from the seam nothing is new: you are inside one formula, so substitute and you are done.
At the seam, the phrase "as x approaches the number" is ambiguous - approaching from the left uses one formula and approaching from the right uses another. So you must ask the two questions separately.
Picture it
Animation
Shows: A seam is where two formulas meet — a rendered Manim animation.
Rendered with Manim.
Takeaway: Each side is evaluated with its own rule. Only then do you compare.
Intuition
Picture two hallways meeting at a doorway. Someone walking up from the left sees the left hallway's floor height. Someone walking up from the right sees the right hallway's.
If both floors arrive at the same height, the doorway is a smooth threshold and there is one answer to "what height are you approaching?"
If the floors arrive at different heights, there is a step. Asking for a single approach height is asking a question with two contradictory answers, so there is no answer at all.
Concept
Deck 1 stated this from the graph. Here is the same rule as a computation you can run.
\[ \lim_{x \to a} f(x) = L \iff \lim_{x \to a^{-}} f(x) = L \ \text{ and } \ \lim_{x \to a^{+}} f(x) = L \]
Both one-sided limits must exist and be the same number. If they disagree, the two-sided limit does not exist, and that is the complete answer.
Notice what is absent from the criterion: the value of the function at the seam. It plays no part whatsoever.
Estimation
Predict first
Find the limit at the seam for this three-line function.
Commit before you compute: what does Worked example: the two sides agree 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 with values on each side of the seam
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 1.99 the first formula gives 3.9601 plus 1, which is 4.9601; at 2.01 the third gives 6.03 minus 1, which is 5.03.
Worked example
Find the limit at the seam for this three-line function.
\[ f(x) = \begin{cases} x^{2} + 1 & x < 2 \\ 10 & x = 2 \\ 3x - 1 & x > 2 \end{cases} \]
Approach from the left using the formula that governs inputs below 2
Why: Every input just below 2 is handled by the first line, so the left-hand limit only sees the squaring formula.
\[ \lim_{x \to 2^{-}} f(x) = (2)^{2} + 1 = 5 \]
Approach from the right using the formula for inputs above 2
Why: Inputs just above 2 are handled by the third line, so the right-hand limit sees the linear formula.
\[ \lim_{x \to 2^{+}} f(x) = 3(2) - 1 = 5 \]
Compare the two and conclude
Why: Both sides deliver 5, so the two-sided limit exists and equals 5 - even though the function was deliberately defined to be 10 at the seam.
\[ \lim_{x \to 2} f(x) = 5 \qquad \text{while} \qquad f(2) = 10 \]
Verify with values on each side of the seam
Why: At 1.99 the first formula gives 3.9601 plus 1, which is 4.9601; at 2.01 the third gives 6.03 minus 1, which is 5.03. Both crowd around 5, and neither is anywhere near 10.
| x | formula used | value |
|---|---|---|
| 1.99 | x squared plus 1 | 4.9601 |
| 1.999 | x squared plus 1 | 4.996001 |
| 2.001 | three x minus 1 | 5.003 |
| 2.01 | three x minus 1 | 5.03 |
Picture it
Animation
Shows: Each line of the worked example "the two sides agree", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 1.99 the first formula gives 3.9601 plus 1, which is 4.9601; at 2.01 the third gives 6.03 minus 1, which is 5.03. Both crowd around 5, and neither is anywhere near 10.
Step zero
Discussion prompt
Worked example: the two sides disagree — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Take the left-hand limit with the linear piece
Answer:
Worked example
Same procedure, different outcome.
\[ g(x) = \begin{cases} x + 3 & x < 1 \\ x^{2} & x \geq 1 \end{cases} \]
Take the left-hand limit with the linear piece
Why: Inputs below 1 use the first line, so the left approach is governed entirely by x plus 3.
\[ \lim_{x \to 1^{-}} g(x) = 1 + 3 = 4 \]
Take the right-hand limit with the squaring piece
Why: Inputs at or above 1 use the second line, so the right approach is governed by x squared.
\[ \lim_{x \to 1^{+}} g(x) = (1)^{2} = 1 \]
Compare: 4 is not 1, so state the failure
Why: The criterion requires the two one-sided limits to agree. They do not, so no single number describes the approach.
\[ \lim_{x \to 1} g(x) \ \text{does not exist (jump discontinuity)} \]
Verify by walking in from both sides
Why: From the left the outputs sit near 4 and from the right they sit near 1. The gap of 3 never closes, no matter how close the inputs get.
| x | value of g |
|---|---|
| 0.99 | 3.99 |
| 0.999 | 3.999 |
| 1.001 | 1.002001 |
| 1.01 | 1.0201 |
Picture it
Animation
Shows: Each line of the worked example "the two sides disagree", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: From the left the outputs sit near 4 and from the right they sit near 1. The gap of 3 never closes, no matter how close the inputs get.
Worked example
This expression has no visible seam, but it has one.
\[ \lim_{x \to 0} \frac{|x|}{x} \]
Rewrite the absolute value as two cases
Why: Absolute value keeps a positive input and flips a negative one, which is exactly a two-line piecewise definition with a seam at zero.
\[ |x| = \begin{cases} x & x > 0 \\ -x & x < 0 \end{cases} \]
Simplify the quotient on each side of the seam
Why: On the positive side the quotient is x over x, which is 1; on the negative side it is negative x over x, which is negative 1.
\[ \lim_{x \to 0^{+}} \frac{|x|}{x} = 1, \qquad \lim_{x \to 0^{-}} \frac{|x|}{x} = -1 \]
Apply the criterion
Why: The one-sided limits are 1 and negative 1. They exist but disagree, so the two-sided limit does not exist.
Verify with real inputs on both sides of zero
Why: The quotient is exactly 1 at every positive input and exactly negative 1 at every negative input, no matter how small. There is nothing to converge to.
| x | absolute value of x | quotient |
|---|---|---|
| -0.1 | 0.1 | -1 |
| -0.001 | 0.001 | -1 |
| 0.001 | 0.001 | 1 |
| 0.1 | 0.1 | 1 |
Picture it
Animation
Shows: Each line of the worked example "an absolute value is a piecewise function in disguise", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The quotient is exactly 1 at every positive input and exactly negative 1 at every negative input, no matter how small. There is nothing to converge to.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The instinct is to find the line that contains the seam and evaluate it.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The value at the seam is what the function IS there.
Ask both one-sided questions, each with its own formula, and only then compare.
Why: The value at the seam is what the function IS there. The limit asks where the outputs are HEADING, which is decided by the neighbours, not the point.
Trap
The instinct is to find the line that contains the seam and evaluate it.
\[ g(x) = \begin{cases} x + 3 & x < 1 \\ x^{2} & x \geq 1 \end{cases} \;\Rightarrow\; \lim_{x \to 1} g(x) \overset{?}{=} g(1) = 1 \]
This answers the wrong question
Why: The value at the seam is what the function IS there. The limit asks where the outputs are HEADING, which is decided by the neighbours, not the point.
The left neighbours were never consulted
Why: Every input just below 1 produces an output near 4, and those inputs are just as close to 1 as the ones on the right. The claimed answer ignores half the picture.
| Side | Formula in force | Value approached |
|---|---|---|
| from the left | x plus 3 | 4 |
| from the right | x squared | 1 |
| claimed limit | used only the point | 1 - wrong |
Ask both one-sided questions, each with its own formula, and only then compare.
\[ \lim_{x \to 1^{-}} g(x) = 4, \qquad \lim_{x \to 1^{+}} g(x) = 1 \]
Compare the two answers before writing anything
Why: The two-sided limit exists only when they match. Here 4 and 1 do not match.
Conclude that the limit does not exist, and note it separately that the value is 1
Why: Both statements are true and independent: g is defined at 1 with value 1, and the limit there fails to exist. A function can be defined at a point where the limit is missing.
Comparison
Comparison matrix
From Trap: reading the seam value instead of the two sides: refill the Formula in force column from what you know. The rest of the table is as it appeared.
| Side | Formula in force | Value approached |
|---|---|---|
| from the left | x plus 3 | 4 |
| from the right | x squared | 1 |
| claimed limit | used only the point | 1 - wrong |
Elimination
Eliminate the wrong options
What is the limit of h 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: From the left, 5 minus 3 gives 2. From the right, 9 minus 4 gives 5. The one-sided limits are 2 and 5, they disagree, so the two-sided limit does not exist.
Check
Compute both one-sided limits with their own formulas before you decide.
\[ h(x) = \begin{cases} 5 - x & x < 3 \\ x^{2} - 4 & x \geq 3 \end{cases} \]
\[ \lim_{x \to 3} h(x) = \; ? \]
Check your understanding
What is the limit of h as x approaches 3?
Answer: A
Why: From the left, 5 minus 3 gives 2. From the right, 9 minus 4 gives 5. The one-sided limits are 2 and 5, they disagree, so the two-sided limit does not exist.
Section
Part 5
Picture it
Figure (svg): Two outer curves closing in on a single common point, with a wiggling curve trapped between them and forced through that point.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Suppose a wandering function is trapped between two others, and those two outer functions both close in on the same height at a point.
Intuition
Suppose a wandering function is trapped between two others, and those two outer functions both close in on the same height at a point.
Figure (svg): Two outer curves closing in on a single common point, with a wiggling curve trapped between them and forced through that point.
The wanderer has no choice. Its room to move shrinks to nothing, so it is forced to the same height. It never needed a formula of its own.
This is the only technique in the deck that computes a limit without simplifying the function. That makes it the tool of last resort for functions that refuse to simplify.
Explain it
Discussion prompt
Explain Trapped between two walls 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:
Suppose a wandering function is trapped between two others, and those two outer functions both close in on the same height at a point.
Concept
Also called the Sandwich Theorem or the Pinching Theorem. Three ingredients, one conclusion.
\[ g(x) \le f(x) \le h(x) \quad \text{for all } x \text{ near } a \text{ (except possibly at } a) \]
\[ \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \;\Longrightarrow\; \lim_{x \to a} f(x) = L \]
Read the hypotheses carefully. The inequality only has to hold near the point, and the two outer limits must be the same number. Miss either requirement and the conclusion is not available.
Analogy
Discussion prompt
Explain The Squeeze Theorem, stated 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:
Also called the Sandwich Theorem or the Pinching Theorem. Three ingredients, one conclusion.
Picture it
Animation
Shows: A region trapped between an upward and a downward parabola meeting at the origin.
The trapped function has no choice.
Takeaway: If the two outer functions agree at a point, anything trapped between them has nowhere else to go — no matter how badly it behaves in between.
Pattern
Predict first
The table runs: 0.1 | 0.01 | -0.0054402 · 0.01 | 0.0001 | -0.0000506
In Worked example: squeezing an oscillator to zero, given the rows so far: what is the next one — the row where x is 0.001?
Correct: 0.001 | 0.000001 | 0.00000083
| x | wall size (x squared) | value of the product |
|---|---|---|
| 0.1 | 0.01 | -0.0054402 |
| 0.01 | 0.0001 | -0.0000506 |
| 0.001 | 0.000001 | 0.00000083 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. The product law needs both factors to have limits.
Worked example
Deck 1's misbehaving function, tamed.
\[ \lim_{x \to 0} x^{2} \sin\!\left(\frac{1}{x}\right) \]
Rule out the product law first
Why: The product law needs both factors to have limits. The sine factor oscillates forever as the input approaches zero and has no limit, so the law does not apply.
Bound the badly behaved factor
Why: Whatever its input is doing, a sine value never leaves the interval from negative one to one. This is the one thing you always know about an oscillating sine.
\[ -1 \le \sin\!\left(\frac{1}{x}\right) \le 1 \quad \text{for all } x \neq 0 \]
Multiply the whole chain by x squared
Why: Multiplying an inequality by a positive quantity preserves its direction, and x squared is positive for every nonzero x, so no inequality flips.
\[ -x^{2} \le x^{2}\sin\!\left(\frac{1}{x}\right) \le x^{2} \]
Take the limit of both outer walls
Why: Both are polynomials, so substitution gives zero for each. The walls close on the same number, which is exactly what the theorem needs.
\[ \lim_{x \to 0} \left(-x^{2}\right) = 0 = \lim_{x \to 0} x^{2} \;\Longrightarrow\; \lim_{x \to 0} x^{2}\sin\!\left(\frac{1}{x}\right) = 0 \]
Verify that real values stay inside the shrinking walls
Why: At 0.1 the product is about negative 0.0054402 with a wall of 0.01; at 0.01 it is about negative 0.0000506 with a wall of 0.0001. Every value is trapped, and the trap is closing on zero.
| x | wall size (x squared) | value of the product |
|---|---|---|
| 0.1 | 0.01 | -0.0054402 |
| 0.01 | 0.0001 | -0.0000506 |
| 0.001 | 0.000001 | 0.00000083 |
Picture it
Animation
Shows: Each line of the worked example "squeezing an oscillator to zero", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 0.1 the product is about negative 0.0054402 with a wall of 0.01; at 0.01 it is about negative 0.0000506 with a wall of 0.0001. Every value is trapped, and the trap is closing on zero.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The tempting generalization: sine is always between negative one and one, so any sine limit must be squeezed.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The theorem requires both walls to approach the SAME number.
Bounded only helps when something else is shrinking the bounds to a point.
Why: The theorem requires both walls to approach the SAME number. These walls sit two units apart forever, so the trapped function has a whole interval to roam in.
Trap
The tempting generalization: sine is always between negative one and one, so any sine limit must be squeezed.
\[ \lim_{x \to 0} \sin\!\left(\frac{1}{x}\right) \;\overset{?}{=}\; 0 \quad \text{"because it is squeezed between } -1 \text{ and } 1\text{"} \]
The two walls here are negative one and one, and they are not closing
Why: The theorem requires both walls to approach the SAME number. These walls sit two units apart forever, so the trapped function has a whole interval to roam in.
The function really does keep hitting both extremes
Why: Arbitrarily close to zero the outputs are exactly 1 and exactly negative 1, again and again. Nothing settles, so the limit does not exist.
| x (about) | the reciprocal (radians) | value of the sine |
|---|---|---|
| 0.6366 | half of pi | 1 |
| 0.2122 | three halves of pi | -1 |
| 0.1273 | five halves of pi | 1 |
| 0.0909 | seven halves of pi | -1 |
Bounded only helps when something else is shrinking the bounds to a point.
\[ -1 \le \sin\!\left(\frac{1}{x}\right) \le 1 \quad \text{gives no conclusion; the walls never meet} \]
Multiply by a factor that goes to zero to force the walls together
Why: The x squared factor crushes the fixed bounds down to zero-width walls, which is what turns a bounded function into a squeezed one.
\[ -x^{2} \le x^{2}\sin\!\left(\frac{1}{x}\right) \le x^{2} \;\Longrightarrow\; \text{limit is } 0 \]
State the rule you actually used
Why: A bounded function times a function going to zero has limit zero. Bounded alone proves nothing; bounded plus a shrinking factor proves everything.
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.
Concept
The Squeeze Theorem's headline result. Substitution gives zero over zero, no factoring helps, and no conjugate helps - but a geometric squeeze on the unit circle settles it.
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad (x \text{ in radians}) \]
The table is not a proof, but it is convincing, and it is worth knowing the numbers.
| x (radians) | sine of x | quotient |
|---|---|---|
| 0.5 | 0.4794255 | 0.9588511 |
| 0.1 | 0.0998334 | 0.9983342 |
| 0.01 | 0.0099998 | 0.9999833 |
| 0.001 | 0.0009999998 | 0.9999998 |
This single fact powers the derivatives of sine and cosine in Deck 9. It is worth memorizing on sight.
Pattern
Step through it
Step through The most important limit in trigonometry one row at a time. What is driving the change, and what would the row after the last one be?
Picture it
Animation
Shows: The most important limit in trigonometry — a rendered Manim animation.
Rendered with Manim.
Takeaway: At zero the formula is undefined, and the curve heads straight for one.
Intuition
On a circle of radius one, an angle measured in radians is the length of the arc it cuts. That is the whole definition of radian measure.
The sine of that angle is the height of the endpoint above the horizontal axis - the straight vertical drop, rather than the curved arc.
For a wide angle the arc is noticeably longer than the height. As the angle shrinks, the arc flattens out until the curve and the straight drop are indistinguishable, so their ratio heads to one.
This is why the formula is false in degrees. In degrees the angle number is not the arc length, and the ratio settles near 0.01745 instead. Every limit and derivative in calculus assumes radians.
Counterexample
Discussion prompt
On a circle of radius one, an angle measured in radians is the length of the arc it cuts. That is the whole definition of radian measure.
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 a wide angle the arc is noticeably longer than the height. As the angle shrinks, the arc flattens out until the curve and the straight drop are indistinguishable, so their ratio heads to one.
Estimation
Predict first
The rule only fires when the quantity inside the sine is identical to the quantity underneath.
Commit before you compute: what does Worked example: making the inside and the bottom match 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 with small inputs on both sides of zero
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 0.001 the sine of 0.005 is about 0.0049999792, and dividing by 0.003 gives about 1.6666597, right beside five thirds, which is about 1.6666667.
Worked example
The rule only fires when the quantity inside the sine is identical to the quantity underneath.
\[ \lim_{x \to 0} \frac{\sin 5x}{3x} \]
Substitute and confirm the indeterminate form
Why: The sine of zero is zero and three times zero is zero, so this is the zero-over-zero form and the rule is a candidate.
Force a five underneath to match the five inside
Why: Multiplying and dividing by 5 changes nothing, but it manufactures the exact pattern the known limit requires.
\[ \frac{\sin 5x}{3x} = \frac{5}{3} \cdot \frac{\sin 5x}{5x} \]
Apply the known limit to the matched piece
Why: As x approaches zero, so does five x, and the sine-over-itself pattern approaches 1 regardless of what the matched quantity is named.
\[ \lim_{x \to 0} \frac{\sin 5x}{3x} = \frac{5}{3} \cdot 1 = \frac{5}{3} \]
Verify with small inputs on both sides of zero
Why: At 0.001 the sine of 0.005 is about 0.0049999792, and dividing by 0.003 gives about 1.6666597, right beside five thirds, which is about 1.6666667.
| x | sine of five x | quotient |
|---|---|---|
| 0.01 | 0.0499792 | 1.6659723 |
| 0.001 | 0.0049999792 | 1.6666597 |
| -0.001 | -0.0049999792 | 1.6666597 |
Picture it
Animation
Shows: Each line of the worked example "making the inside and the bottom match", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 0.001 the sine of 0.005 is about 0.0049999792, and dividing by 0.003 gives about 1.6666597, right beside five thirds, which is about 1.6666667.
Step zero
Discussion prompt
Worked example: a sine over a sine — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Divide the top and the bottom by x
Answer:
Worked example
Two sines, two different multipliers. Split it into two copies of the known pattern.
\[ \lim_{x \to 0} \frac{\sin 4x}{\sin 7x} \]
Divide the top and the bottom by x
Why: Dividing both parts of a fraction by the same nonzero quantity leaves the fraction alone, and it sets up a sine-over-its-own-input pattern in each part.
\[ \frac{\sin 4x}{\sin 7x} = \frac{\dfrac{\sin 4x}{x}}{\dfrac{\sin 7x}{x}} \]
Match each numerator with its own inside quantity
Why: Write the top as 4 times sine of four x over four x, and the bottom as 7 times sine of seven x over seven x. Each fraction is now exactly the known pattern.
\[ = \frac{4 \cdot \dfrac{\sin 4x}{4x}}{7 \cdot \dfrac{\sin 7x}{7x}} \;\longrightarrow\; \frac{4 \cdot 1}{7 \cdot 1} \]
Use the quotient law on the two matched pieces
Why: Each piece now has a limit and the bottom limit is 7, which is not zero, so the quotient law finally applies legally.
\[ \lim_{x \to 0} \frac{\sin 4x}{\sin 7x} = \frac{4}{7} \]
Verify at a small input
Why: At 0.001 the sine of 0.004 is about 0.0039999893 and the sine of 0.007 is about 0.0069999428; their ratio is about 0.5714317, and four sevenths is about 0.5714286.
| x | sine of four x | sine of seven x | ratio |
|---|---|---|---|
| 0.001 | 0.0039999893 | 0.0069999428 | 0.5714317 |
Concept
The sine limit has a partner that shows up just as often, and the conjugate trick from Part 3 proves it.
\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \]
Multiply top and bottom by the conjugate of the numerator, then use the Pythagorean identity.
\[ \frac{1-\cos x}{x}\cdot\frac{1+\cos x}{1+\cos x} = \frac{1-\cos^{2}x}{x(1+\cos x)} = \frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x} \]
\[ \longrightarrow \; 1 \cdot \frac{0}{2} = 0 \]
Notice the payoff: three separate techniques from this deck - the conjugate, the sine limit, and the product law - cooperating in one computation.
Picture it
Animation
Shows: The companion limit for cosine — a rendered Manim animation.
Rendered with Manim.
Takeaway: The conjugate trick again, in a trigonometric costume.
Ranking
Put in order
Put the moves of Worked example: tangent over its angle 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. That is the definition, and it converts an unfamiliar limit into the one pattern you already know plus a harmless extra factor.
Worked example
There is no separate tangent rule to memorize. Break the tangent into its definition.
\[ \lim_{x \to 0} \frac{\tan x}{x} \]
Rewrite the tangent as sine over cosine
Why: That is the definition, and it converts an unfamiliar limit into the one pattern you already know plus a harmless extra factor.
\[ \frac{\tan x}{x} = \frac{\sin x}{x \cos x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x} \]
Take the limit of each factor separately
Why: The first factor is the known sine limit, equal to 1. The second is continuous at zero with cosine of zero equal to 1, so substitution handles it.
\[ \lim_{x \to 0} \frac{\tan x}{x} = 1 \cdot \frac{1}{1} = 1 \]
Check that the product law was legal here
Why: Both factors have real limits, 1 and 1, so the product law applies. It is worth naming the law rather than just multiplying on instinct.
Verify numerically near zero
Why: At 0.1 the tangent is about 0.1003347 and the quotient is about 1.003347; at 0.01 the quotient is about 1.0000333. The values approach 1 from above, which fits since the tangent slightly exceeds its angle.
| x | tangent of x | quotient |
|---|---|---|
| 0.1 | 0.1003347 | 1.003347 |
| 0.01 | 0.0100003 | 1.0000333 |
Picture it
Animation
Shows: Each line of the worked example "tangent over its angle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both factors have real limits, 1 and 1, so the product law applies. It is worth naming the law rather than just multiplying on instinct.
Check
You are told nothing about the formula for the function - only that it is trapped.
\[ 4x - 9 \;\le\; f(x) \;\le\; x^{2} - 4x + 7 \quad \text{for all } x \]
\[ \lim_{x \to 4} f(x) = \; ? \]
Check your understanding
What does the Squeeze Theorem give for this limit?
Answer: A
Why: At 4 the lower bound is 16 minus 9, which is 7, and the upper bound is 16 minus 16 plus 7, which is also 7. Both walls close on 7, so the trapped function is forced to 7.
Section
Part 6
Constraint
Discussion prompt
Run Pattern: the master procedure for any limit with this step confiscated:
A nonzero number over zero? This is not indeterminate. Examine each side; expect an infinite limit or a two-sided failure, and name the vertical asymptote.
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
Every limit in this deck, and most on your first exam, is handled by this sequence. Run it in order and never skip step one.
A limit at a seam or an absolute value is the one detour: go straight to the two one-sided limits, then compare.
Edge cases
Discussion prompt
Pattern: the master procedure for any limit 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:
Every limit in this deck, and most on your first exam, is handled by this sequence. Run it in order and never skip step one.
Pattern
Once substitution has given you the indeterminate form, the shape of the expression tells you which tool to reach for. There are only five.
| What you see | Tool | The move |
|---|---|---|
| Two polynomials | Factor and cancel | Factor out the linear factor for the number you are approaching, then divide it out. |
| A square root in a difference | Conjugate | Multiply top and bottom by the same two terms with the middle sign flipped. |
| A fraction inside a fraction | Clear the complex fraction | Combine the inner fractions over one denominator, then multiply by the reciprocal. |
| A seam or an absolute value | One-sided limits | Compute each side with its own formula and compare them. |
| A sine or cosine heading to zero | The trig limits | Match the inside quantity to the denominator, or use the Squeeze Theorem. |
When two tools both look plausible, prefer the one that leaves less algebra. A quotient you can split apart by hand rarely needs a conjugate.
Trade off
Comparison matrix
From Pattern: pick the tool from the shape: every row here is a choice with a cost. Fill the Tool column, then say which row you would actually pick and what you give up for it.
| What you see | Tool | The move |
|---|---|---|
| Two polynomials | Factor and cancel | Factor out the linear factor for the number you are approaching, then divide it out. |
| A square root in a difference | Conjugate | Multiply top and bottom by the same two terms with the middle sign flipped. |
| A fraction inside a fraction | Clear the complex fraction | Combine the inner fractions over one denominator, then multiply by the reciprocal. |
| A seam or an absolute value | One-sided limits | Compute each side with its own formula and compare them. |
| A sine or cosine heading to zero | The trig limits | Match the inside quantity to the denominator, or use the Squeeze Theorem. |
Elimination
Eliminate the wrong options
What is the correct first move, and what does the limit turn out to be?
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 conjugate makes the numerator x minus 9, which cancels the denominator and leaves 1 over the square root of x plus 3. At 9 that is 1 over 6.
Check
Substitution gives the indeterminate form here. What is the correct next move?
\[ \lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9} \]
Check your understanding
What is the correct first move, and what does the limit turn out to be?
Answer: A
Why: The conjugate makes the numerator x minus 9, which cancels the denominator and leaves 1 over the square root of x plus 3. At 9 that is 1 over 6.
Check
Run the whole procedure: substitute, identify the form, choose the tool, simplify, substitute again.
\[ \lim_{x \to 1} \frac{x^{2} + 2x - 3}{x^{2} - 1} \]
Check your understanding
What is the value of this limit?
Answer: A
Why: Both parts vanish at 1. The top factors as x plus 3 times x minus 1 and the bottom as x plus 1 times x minus 1. After cancelling, x plus 3 over x plus 1 at 1 gives 4 over 2, which is 2.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — The Limit Laws · The Zero-Over-Zero Form · Conjugates and Complex Fractions · Piecewise Functions at the Seam · The Squeeze Theorem · Putting It Together. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You started this deck able to see a limit. You can now compute one exactly, and you know why each move is allowed.
Two ideas to carry into the next deck. First, substitution works precisely when the graph has no break at the point - that property is called continuity, and Deck 3 makes it official. Second, the zero-over-zero form you spent this deck defusing is exactly the form every derivative is built from, so this algebra is about to become daily work.
| Trigger | Response |
|---|---|
| Substitution gives a number | That is the limit. |
| Nonzero over zero | Check both sides; expect a vertical asymptote. |
| Zero over zero | Factor, conjugate, or clear the complex fraction. |
| A seam or an absolute value | Two one-sided limits, then compare. |
| Bounded times something shrinking | Squeeze it. |
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