This deck is about the end behavior of functions. It explains what a limit at infinity means, then covers the divide-by-the-highest-power technique for rational functions and the three degree cases, horizontal and slant asymptotes, radicals in the negative direction, and how exponentials outrun polynomials. It targets the sign error that appears when you pull a variable out of a square root, the myth that a graph never crosses its horizontal asymptote, and the confusion between vertical-asymptote limits and limits at infinity.
Subject: Calculus I · 136 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 04
End behavior, horizontal and slant asymptotes, and the sign trap inside a square root.
Objectives
This deck is about the far ends of a graph - what a function settles down to when the input runs away forever.
Warm-up
Discussion prompt
Before we open Limits at Infinity, End Behavior, and Asymptotes: without looking back, what was the main idea of Continuity and the Intermediate Value Theorem, 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:
That deck gives the three-part definition of continuity at a point and uses it to classify removable, jump, and infinite discontinuities. It then covers continuity on intervals and for the standard function families, composites and passing a limit inside, and choosing parameters that make a piecewise function continuous, before closing with the Intermediate Value Theorem and bisection. It targets the students who check the limit but forget that the function must be defined, who think cancelling a factor erases the hole, and who use the IVT without checking continuity or read it as guaranteeing exactly one root.
Section
Section 1
Concept
Every limit you have taken so far asked: as the input closes in on one particular number, where does the output head?
A limit at infinity asks something different: as the input runs to the right forever, does the output settle down to a number?
\[ \lim_{x \to \infty} f(x) = L \]
Read it as: the outputs get and stay as close to that number as you like, once the input is large enough.
Counterexample
Discussion prompt
Every limit you have taken so far asked: as the input closes in on one particular number, where does the output head?
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:
A limit at infinity asks something different: as the input runs to the right forever, does the output settle down to a number?
Intuition
Picture yourself walking to the right along the graph. You never stop. You are not aiming at a spot on the curve - you are watching your altitude.
If your altitude keeps drifting toward the same height and stays there, that height is the limit at infinity. If it keeps climbing, or keeps bouncing, there is no such height.
That is the whole idea. Everything else in this deck is a technique for finding the height without walking.
Analogy
Discussion prompt
Explain Walk to the right forever and look at the height 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:
Picture yourself walking to the right along the graph. You never stop. You are not aiming at a spot on the curve - you are watching your altitude.
Concept
limit at infinity — A description of end behavior. The infinity symbol underneath the limit is shorthand for 'let the input grow without bound' - it is never a value you substitute.
There are two ends, so there are two questions, and their answers can differ.
\[ \lim_{x \to \infty} f(x) \qquad \text{and} \qquad \lim_{x \to -\infty} f(x) \]
Always ask which end you are on. Half the errors in this topic come from answering the wrong question.
Explain it
Discussion prompt
Explain The symbol is not a number 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:
Always ask which end you are on. Half the errors in this topic come from answering the wrong question.
Concept
Divide one by a bigger and bigger number and the result shrinks toward nothing. That is the engine of this entire deck.
\[ \lim_{x \to \infty} \frac{1}{x} = 0 \qquad \lim_{x \to -\infty} \frac{1}{x} = 0 \]
At the left end the outputs are negative, but they still shrink toward zero. Approaching zero from below is still approaching zero.
Picture it
Animation
Shows: The one fact everything is built on — a rendered Manim animation.
Rendered with Manim.
Takeaway: Any positive power in the denominator drives the whole thing to zero.
Ranking
Put in order
Put the moves of Worked example: watch it happen numerically 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. Powers of ten make the pattern visible in one glance instead of ten.
Worked example
Before trusting a rule, look at the numbers. Take the reciprocal of the input and let the input grow.
\[ f(x) = \frac{1}{x} \]
Evaluate at inputs that grow by factors of ten
Why: Powers of ten make the pattern visible in one glance instead of ten.
| input | output |
|---|---|
| 1 | 1 |
| 10 | 0.1 |
| 100 | 0.01 |
| 1000 | 0.001 |
| 1000000 | 0.000001 |
Read the trend
Why: Each tenfold jump in the input cuts the output to a tenth. Nothing stops that, so the outputs have no floor above zero.
Do the same at the left end
Why: At the negative inputs the outputs are negative but equally tiny, so the drift is toward zero from below.
| input | output |
|---|---|
| -10 | -0.1 |
| -100 | -0.01 |
| -1000 | -0.001 |
Verify the claim against the definition
Why: To be within one thousandth of zero we only need the input past one thousand, and every larger input stays inside. That is exactly what the limit statement promises.
\[ \lim_{x \to \pm\infty} \frac{1}{x} = 0 \]
Picture it
Animation
Shows: Each line of the worked example "watch it happen numerically", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: To be within one thousandth of zero we only need the input past one thousand, and every larger input stays inside. That is exactly what the limit statement promises.
Concept
If the reciprocal of the input dies, so does the reciprocal of any positive power of it - faster, in fact.
\[ \lim_{x \to \infty} \frac{1}{x^{p}} = 0 \quad \text{for every } p > 0 \]
This covers square roots too, since a square root is a fractional power. Keep this fact in your pocket - the next technique manufactures these terms on purpose.
\[ \lim_{x \to \infty} \frac{1}{\sqrt{x}} = 0, \qquad \lim_{x \to \infty} \frac{1}{x^{3}} = 0 \]
Concept
At the left end the input is negative, so an odd power stays negative and an even power turns positive.
\[ \lim_{x \to -\infty} \frac{1}{x^{3}} = 0^{-}, \qquad \lim_{x \to -\infty} \frac{1}{x^{2}} = 0^{+} \]
The limit is zero either way. The sign only matters when this term sits inside a larger expression whose sign you are tracking.
Concept
horizontal asymptote — A horizontal line that the graph approaches as the input runs to positive or negative infinity. It is the geometric picture of a finite limit at infinity - nothing more.
\[ \lim_{x \to \infty} f(x) = L \quad \Longleftrightarrow \quad y = L \text{ is a horizontal asymptote} \]
So finding a horizontal asymptote is not a separate skill. It is the same limit, reported as a line instead of a number.
Definition probe
Sort into buckets
Every line below is part of the definition of limit at infinity or of horizontal asymptote — one or the other, never both. Put each where it belongs.
Concept
Figure (svg): A curve rising from the lower left and flattening out as it approaches a dashed horizontal line labelled y equals L
The dashed line is the asymptote. The curve is not required to reach it, and - as you will see shortly - it is not forbidden from touching it either.
The only promise is about the gap: it shrinks to nothing as you go right.
A function can have a different horizontal asymptote at each end, one at only one end, or none at all.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Asked for the end behavior of this function, a student sees the value that breaks the denominator and answers from that.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The word 'asymptote' triggered the wrong picture.
Read the arrow under the limit first. It tells you which question is being asked.
Why: The word 'asymptote' triggered the wrong picture. The blow-up happens near the input value two, which has nothing to do with the far right.
Trap
Asked for the end behavior of this function, a student sees the value that breaks the denominator and answers from that.
\[ f(x) = \frac{1}{x-2} \]
Answer: the limit at infinity grows without bound, because the graph blows up at the asymptote
Why: The word 'asymptote' triggered the wrong picture. The blow-up happens near the input value two, which has nothing to do with the far right.
\[ \lim_{x \to \infty} \frac{1}{x-2} = \infty \quad \text{(wrong)} \]
Read the arrow under the limit first. It tells you which question is being asked.
\[ f(x) = \frac{1}{x-2} \]
The arrow points to infinity, so let the input grow
Why: For huge inputs the denominator is huge, so the fraction is tiny. The end behavior is a flat line at height zero.
\[ \lim_{x \to \infty} \frac{1}{x-2} = 0 \]
| input | output |
|---|---|
| 12 | 0.1 |
| 102 | 0.01 |
| 1002 | 0.001 |
The blow-up is a separate, one-sided question at the input two
Why: Both facts are true about the same function; they describe different parts of the picture. One gives a vertical asymptote, the other a horizontal one.
\[ \lim_{x \to 2^{+}} \frac{1}{x-2} = \infty, \qquad \lim_{x \to 2^{-}} \frac{1}{x-2} = -\infty \]
Pattern
Step through it
Step through Trap: two completely different questions one row at a time. What is driving the change, and what would the row after the last one be?
Section
Section 2
Intuition
Suppose a company's revenue is a million dollars plus five dollars per customer. At ten customers the million dominates completely.
Now flip it: at ten million customers, the five-dollar term is worth fifty million and the fixed million is a rounding error.
Polynomials behave the same way. At the far ends the highest-degree term swamps everything else, and the lower terms become noise.
\[ 3x^2 + 5x - 1 \;\approx\; 3x^2 \quad \text{when } x \text{ is huge} \]
So end behavior of a rational function is really a contest between two leading terms. The technique below just makes that contest visible.
Picture it
Animation
Shows: A rational function flattening toward a horizontal asymptote at both ends.
The small terms lose, eventually.
Takeaway: Far from the origin the lower-order terms stop mattering, and the curve settles toward the ratio of the leading coefficients.
Concept
You cannot substitute, because both top and bottom run away. So change the form of the fraction without changing its value.
Find the largest power of the variable in the denominator, then divide every single term - top and bottom - by it.
\[ \frac{3x^2 + 5x - 1}{2x^2 - 7} \;=\; \frac{\dfrac{3x^2}{x^2} + \dfrac{5x}{x^2} - \dfrac{1}{x^2}}{\dfrac{2x^2}{x^2} - \dfrac{7}{x^2}} \]
Dividing top and bottom by the same nonzero quantity leaves the function alone. What it buys you is a pile of reciprocal terms, and you already know every one of those dies.
Picture it
Animation
Shows: The technique: divide by the highest power — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every leftover term carries a power of x underneath, and dies.
Step zero
Discussion prompt
Worked example: equal degrees — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Notice substitution fails
Answer:
Worked example
Find the end behavior at the right end.
\[ \lim_{x \to \infty} \frac{3x^2 + 5x - 1}{2x^2 - 7} \]
Notice substitution fails
Why: Both the numerator and the denominator grow without bound, so this is the indeterminate contest, not an answer.
The highest power in the denominator is the square, so divide every term by it
Why: Same quantity top and bottom means the function is unchanged, but now every leftover term is a reciprocal power.
\[ = \lim_{x \to \infty} \frac{3 + \dfrac{5}{x} - \dfrac{1}{x^2}}{2 - \dfrac{7}{x^2}} \]
Send each reciprocal term to zero
Why: Every term with the variable in a denominator vanishes at infinity. Only the constants survive.
\[ = \frac{3 + 0 - 0}{2 - 0} = \frac{3}{2} \]
State the answer as a limit and as a line
Why: A finite limit at infinity is exactly a horizontal asymptote, so both forms are the same fact.
\[ \lim_{x \to \infty} \frac{3x^2+5x-1}{2x^2-7} = \frac{3}{2}, \qquad y = \tfrac{3}{2} \text{ is a horizontal asymptote} \]
Verify with a table of large inputs
Why: The computed values close in on 1.5, which confirms the algebra rather than merely restating it.
| input | value of the fraction |
|---|---|
| 100 | 1.52548 |
| 1000 | 1.50250 |
| 10000 | 1.50025 |
Picture it
Animation
Shows: Each line of the worked example "equal degrees", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The computed values close in on 1.5, which confirms the algebra rather than merely restating it.
Ranking
Put in order
These are the steps of The recipe for any rational end behavior, 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
This works for every rational function, at either end, every time.
Step two is the one students get wrong. It is the highest power in the bottom, not the top - that choice is what forces the denominator to settle to a nonzero constant.
Elimination
Eliminate the wrong options
What is the limit of (6x^2 - x + 4)/(3x^2 + 5x) as x grows without bound?
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: Divide every term by x squared to get (6 - 1/x + 4/x^2) over (3 + 5/x). Every reciprocal term dies, leaving 6/3 = 2. Numeric check at x = 1000: the fraction equals 5999004/3005000, about 1.9963.
Check
Work it on paper first. Divide through by the highest power in the denominator.
\[ \lim_{x \to \infty} \frac{6x^2 - x + 4}{3x^2 + 5x} \]
Check your understanding
What is the limit of (6x^2 - x + 4)/(3x^2 + 5x) as x grows without bound?
Answer: A
Why: Divide every term by x squared to get (6 - 1/x + 4/x^2) over (3 + 5/x). Every reciprocal term dies, leaving 6/3 = 2. Numeric check at x = 1000: the fraction equals 5999004/3005000, about 1.9963.
Concept
Once you have done the division a few times, you can see the outcome from the degrees alone. There are exactly three ways the contest ends.
Bottom-heavy. The denominator has the larger degree, so the bottom outruns the top and the fraction collapses to zero.
Equal degrees. Neither side outruns the other; the limit is the ratio of the two leading coefficients.
Top-heavy. The numerator has the larger degree, so the fraction grows without bound. There is no horizontal asymptote at that end.
Picture it
Animation
Shows: The three degree comparisons and the horizontal asymptote each produces.
Compare degrees, then stop.
Takeaway: Compare the degrees and the answer is already determined: zero, the coefficient ratio, or no horizontal asymptote at all.
Pattern
Predict first
The table runs: 10 | 0.48454 · 100 | 0.04071
In Worked example: bottom-heavy, given the rows so far: what is the next one — the row where input is 1000?
Correct: 1000 | 0.00401
| input | value of the fraction |
|---|---|
| 10 | 0.48454 |
| 100 | 0.04071 |
| 1000 | 0.00401 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. Follow the recipe even when you can guess the answer - it is what keeps the guess honest.
Worked example
\[ \lim_{x \to \infty} \frac{4x + 7}{x^2 - 3} \]
The highest power in the denominator is the square, so divide every term by it
Why: Follow the recipe even when you can guess the answer - it is what keeps the guess honest.
\[ = \lim_{x \to \infty} \frac{\dfrac{4}{x} + \dfrac{7}{x^2}}{1 - \dfrac{3}{x^2}} \]
Send every reciprocal term to zero
Why: The whole numerator was made of reciprocal terms, so the top collapses while the bottom keeps its 1.
\[ = \frac{0 + 0}{1 - 0} = 0 \]
Name the asymptote
Why: A limit of zero at infinity is the horizontal line at height zero, which is the x-axis.
\[ y = 0 \]
Verify numerically
Why: The values shrink by roughly a factor of ten each time the input does, exactly as a linear-over-quadratic fraction should.
| input | value of the fraction |
|---|---|
| 10 | 0.48454 |
| 100 | 0.04071 |
| 1000 | 0.00401 |
Picture it
Animation
Shows: Each line of the worked example "bottom-heavy", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The values shrink by roughly a factor of ten each time the input does, exactly as a linear-over-quadratic fraction should.
Pattern
Predict first
The table runs: 10 | 18.95 | 20 · 100 | 199.89 | 200
In Worked example: top-heavy, given the rows so far: what is the next one — the row where input is 1000?
Correct: 1000 | 1999.99 | 2000
| input | value of the fraction | twice the input |
|---|---|---|
| 10 | 18.95 | 20 |
| 100 | 199.89 | 200 |
| 1000 | 1999.99 | 2000 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. Dividing by the top's cube instead would send the denominator to zero and hide the answer, so stay with the denominator's power.
Worked example
\[ \lim_{x \to \infty} \frac{2x^3 - x}{x^2 + 5} \]
Divide by the highest power in the denominator, which is the square
Why: Dividing by the top's cube instead would send the denominator to zero and hide the answer, so stay with the denominator's power.
\[ = \lim_{x \to \infty} \frac{2x - \dfrac{1}{x}}{1 + \dfrac{5}{x^2}} \]
Read the surviving pieces
Why: The reciprocal terms die, but a bare linear term is left upstairs. The denominator settles on 1, so the fraction behaves like twice the input.
\[ \approx \frac{2x}{1} = 2x \]
Conclude that the values grow without bound
Why: Twice a runaway input is still a runaway. There is no finite height, so there is no horizontal asymptote at this end.
\[ \lim_{x \to \infty} \frac{2x^3 - x}{x^2+5} = \infty \]
Verify that the values track twice the input
Why: Each computed value sits just under twice its input, which is exactly what the leftover linear term predicted.
| input | value of the fraction | twice the input |
|---|---|---|
| 10 | 18.95 | 20 |
| 100 | 199.89 | 200 |
| 1000 | 1999.99 | 2000 |
Picture it
Animation
Shows: Each line of the worked example "top-heavy", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Each computed value sits just under twice its input, which is exactly what the leftover linear term predicted.
Concept
Writing that a limit equals infinity is a description of how it fails, not a value. Infinity is not a number the function reaches.
Saying it grows without bound is more informative than saying the limit does not exist, because it tells you the direction. But if a test asks whether the limit exists, the answer is no.
Both statements below are correct and they are not in conflict.
\[ \lim_{x \to \infty} \frac{2x^3-x}{x^2+5} = \infty \quad \text{and} \quad \text{the limit does not exist} \]
Concept
Compare the degree of the top with the degree of the bottom.
| Degrees | Limit at either end | Horizontal asymptote |
|---|---|---|
| top degree less than bottom degree | 0 | the x-axis |
| top degree equals bottom degree | ratio of leading coefficients | that ratio |
| top degree greater than bottom degree | grows without bound | none |
Use this to predict, then run the division to confirm. The shortcut is safe for the size of the answer, but it hides one thing: the sign in the top-heavy case.
Comparison
Comparison matrix
From The three cases on one card: refill the Limit at either end column from what you know. The rest of the table is as it appeared.
| Degrees | Limit at either end | Horizontal asymptote |
|---|---|---|
| top degree less than bottom degree | 0 | the x-axis |
| top degree equals bottom degree | ratio of leading coefficients | that ratio |
| top degree greater than bottom degree | grows without bound | none |
Anomaly
Predict first
A student writes this, and it looks reasonable:
The task is the left end of a top-heavy fraction.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The degree comparison was correct, but it only reports the size of the answer.
Divide through and look at what actually survives, sign included.
Why: The degree comparison was correct, but it only reports the size of the answer. The sign was never checked.
Trap
The task is the left end of a top-heavy fraction.
\[ \lim_{x \to -\infty} \frac{2x^3 - x}{5x^2 + 1} \]
Answer: the top has the bigger degree, so the limit is infinity
Why: The degree comparison was correct, but it only reports the size of the answer. The sign was never checked.
\[ \lim_{x \to -\infty} \frac{2x^3-x}{5x^2+1} = \infty \quad \text{(wrong)} \]
Divide through and look at what actually survives, sign included.
\[ \lim_{x \to -\infty} \frac{2x^3 - x}{5x^2 + 1} \]
Divide every term by the square
Why: The denominator's highest power is the square, so this is the recipe's step two.
\[ = \lim_{x \to -\infty} \frac{2x - \dfrac{1}{x}}{5 + \dfrac{1}{x^2}} \;\approx\; \frac{2x}{5} \]
Now put the direction in
Why: Two fifths of a large negative number is a large negative number, so the values run down, not up.
\[ \lim_{x \to -\infty} \frac{2x^3-x}{5x^2+1} = -\infty \]
| input | value of the fraction |
|---|---|
| -100 | -39.997 |
| -1000 | -400.00 |
Trade off
Comparison matrix
From Trap: top-heavy does not mean positive: every row here is a choice with a cost. Fill the value of the fraction column, then say which row you would actually pick and what you give up for it.
| input | value of the fraction |
|---|---|
| -100 | -39.997 |
| -1000 | -400.00 |
Step zero
Discussion prompt
Worked example: opposite ends, opposite answers — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Divide every term by the square
Answer:
Worked example
One function, two questions. Notice how little of the work changes and how much of the answer does.
\[ g(x) = \frac{-5x^3 + 2}{x^2 + 1} \]
Divide every term by the square
Why: The denominator's highest power is the square; the recipe does not care which end you are headed for.
\[ g(x) = \frac{-5x + \dfrac{2}{x^2}}{1 + \dfrac{1}{x^2}} \;\approx\; -5x \]
Right end: the input is large and positive
Why: Negative five times a large positive number is a large negative number.
\[ \lim_{x \to \infty} g(x) = -\infty \]
Left end: the input is large and negative
Why: Negative five times a large negative number is a large positive number. The negative leading coefficient flips the answer between the ends.
\[ \lim_{x \to -\infty} g(x) = \infty \]
Verify both ends numerically
Why: Each value sits within a hundredth of a percent of negative five times the input, and the two ends genuinely have opposite signs.
| input | value of g | negative five times the input |
|---|---|---|
| 100 | -499.95 | -500 |
| -100 | 499.95 | 500 |
Picture it
Animation
Shows: Each line of the worked example "opposite ends, opposite answers", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Each value sits within a hundredth of a percent of negative five times the input, and the two ends genuinely have opposite signs.
Prediction
Predict first
What is the limit of (7x^3 + 2x)/(4 - x^2) as x runs to negative infinity?
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 grows without bound in the positive direction
Why: Dividing by x squared leaves (7x + 2/x) over (4/x^2 - 1). The top runs to negative infinity and the bottom tends to -1, so the quotient runs to positive infinity. Spot-check at x = -100: the value is -7000200/-9996, about +700.3.
Check
Watch the leading coefficient in the denominator here - it is hiding a minus sign.
\[ \lim_{x \to -\infty} \frac{7x^3 + 2x}{4 - x^2} \]
Check your understanding
What is the limit of (7x^3 + 2x)/(4 - x^2) as x runs to negative infinity?
Answer: A
Why: Dividing by x squared leaves (7x + 2/x) over (4/x^2 - 1). The top runs to negative infinity and the bottom tends to -1, so the quotient runs to positive infinity. Spot-check at x = -100: the value is -7000200/-9996, about +700.3.
Section
Section 3
Concept
There are only two ends, so there are at most two answers - one for the right end and one for the left.
For a rational function the two ends always agree, because the leading terms do not change. For other functions they often disagree.
\[ \lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}} = 1, \qquad \lim_{x \to -\infty} \frac{x}{\sqrt{x^2+1}} = -1 \]
That function has two horizontal asymptotes, one at each end. You will see exactly why the left end flips sign in the next section.
Concept
A vertical asymptote is a wall - the function is undefined there, so the graph genuinely cannot touch it.
A horizontal asymptote is not a wall. It is only a promise about the tail. Anything can happen before the tail, including crossing the line, touching it, or crossing it infinitely often.
The classic example crosses its asymptote at the x-axis forever, once between each pair of consecutive multiples of the half-turn.
\[ f(x) = \frac{\sin x}{x}, \qquad \lim_{x \to \infty} f(x) = 0 \]
So never answer a crossing question by instinct. Answer it by solving an equation.
Picture it
Animation
Shows: A decaying oscillation crossing its horizontal asymptote repeatedly.
An asymptote is not a fence.
Takeaway: An asymptote describes the far-away behaviour, not a barrier. This curve crosses it infinitely often and still converges to it.
Estimation
Predict first
Find the horizontal asymptote, then find every point where the graph meets it.
Commit before you compute: what does Worked example: where does it cross the asymptote? 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 by substituting back and by checking the tail
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. The point really is on the line, and far out on the right the values sit just under three, so the graph crosses once and then settles from below.
Worked example
\[ f(x) = \frac{3x^2 - 6x}{x^2 + 2} \]
Find the horizontal asymptote, then find every point where the graph meets it.
Find the asymptote from the degrees
Why: The degrees are equal, so the limit at either end is the ratio of the leading coefficients, three over one.
\[ y = 3 \]
To meet the line, the output must actually equal three
Why: Crossing is not a limit question at all; it is the ordinary algebra question of solving an equation.
\[ \frac{3x^2 - 6x}{x^2+2} = 3 \]
Clear the denominator and simplify
Why: The denominator is never zero, so multiplying through is safe and introduces no false solutions.
\[ 3x^2 - 6x = 3x^2 + 6 \;\Longrightarrow\; -6x = 6 \]
Solve
Why: The squared terms cancel, leaving a linear equation with a single solution.
\[ x = -1 \]
Verify by substituting back and by checking the tail
Why: The point really is on the line, and far out on the right the values sit just under three, so the graph crosses once and then settles from below.
\[ f(-1) = \frac{3 + 6}{1 + 2} = \frac{9}{3} = 3 \quad \checkmark \]
| input | value of f |
|---|---|
| -1 | 3.0000 |
| 100 | 2.9394 |
| -100 | 3.0594 |
Picture it
Animation
Shows: Each line of the worked example "where does it cross the asymptote?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The point really is on the line, and far out on the right the values sit just under three, so the graph crosses once and then settles from below.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The task: does this graph ever meet its horizontal asymptote?
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This is a memory of vertical asymptotes leaking into a horizontal one.
Find the asymptote, then solve for equality like any other equation.
Why: This is a memory of vertical asymptotes leaking into a horizontal one. Nothing about a limit at infinity forbids the value from being hit.
Trap
The task: does this graph ever meet its horizontal asymptote?
\[ h(x) = \frac{2x + 4}{x^2 + 1} \]
Answer: the asymptote is the x-axis, and a graph can never touch its asymptote
Why: This is a memory of vertical asymptotes leaking into a horizontal one. Nothing about a limit at infinity forbids the value from being hit.
Conclusion recorded: no crossing points.
Find the asymptote, then solve for equality like any other equation.
\[ h(x) = \frac{2x + 4}{x^2 + 1} \]
The bottom has the larger degree, so the asymptote is the x-axis
Why: Bottom-heavy means the fraction collapses to zero at both ends.
\[ y = 0 \]
A fraction is zero exactly when its numerator is zero
Why: The denominator is never zero, so the only condition left is on the top.
\[ 2x + 4 = 0 \;\Longrightarrow\; x = -2 \]
Verify the point sits on the line
Why: The graph lands exactly on its asymptote at that input, then leaves and approaches it forever from the other side.
\[ h(-2) = \frac{-4+4}{4+1} = \frac{0}{5} = 0 \quad \checkmark \]
Notation
Annotate
From Trap: 'it approaches but never touches' — read this one piece at a time. What is each part doing?
On: \( h(x) = \frac{2x + 4}{x^2 + 1} \)
Check
Both parts are degree one. Watch the sign on the leading coefficient of the numerator.
\[ f(x) = \frac{5 - 2x}{3x + 7} \]
Check your understanding
What is the horizontal asymptote of f(x) = (5 - 2x)/(3x + 7)?
Answer: A
Why: The degrees match, so the limit is the ratio of leading coefficients: -2 over 3. Numeric check at x = 1000: the value is -1995/3007, about -0.6635, closing in on -0.6667.
Section
Section 4
Concept
A square root always returns the nonnegative option. So squaring and then rooting does not give you back what you started with when you started negative.
\[ \sqrt{(-7)^2} = \sqrt{49} = 7 \neq -7 \]
the root-of-a-square identity — The square root of a quantity squared equals the absolute value of that quantity - never the quantity itself, unless you already know it is nonnegative.
\[ \sqrt{x^2} = |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases} \]
This single identity is responsible for almost every sign error in end-behavior problems with radicals.
Picture it
Animation
Shows: A square root of a square is an absolute value — a rendered Manim animation.
Rendered with Manim.
Takeaway: This single line is why the left end so often flips sign.
Intuition
To divide through by the input, you have to push it inside the square root. Pushing a quantity inside a root means squaring it first.
Squaring destroys the sign. The absolute value is the receipt that records the sign you destroyed, so you can put it back.
At the right end the input is positive, so the receipt says nothing changed. At the left end it says a minus sign is owed - and that minus sign is the whole answer.
Practical rule: before you touch a radical end-behavior problem, write down whether the input is positive or negative. Everything follows from that one line.
Hypothesis
Predict first
Worked example: a radical at the right end 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: Record the direction: the input is large and positive
Why: This is the line that decides the sign later, so it goes down before any algebra.
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
\[ \lim_{x \to \infty} \frac{\sqrt{4x^2 + 1}}{x + 3} \]
Record the direction: the input is large and positive
Why: This is the line that decides the sign later, so it goes down before any algebra.
Divide top and bottom by the input
Why: The denominator's highest power is the first power, so that is what the recipe divides by.
\[ = \lim_{x \to \infty} \frac{\dfrac{\sqrt{4x^2+1}}{x}}{\dfrac{x+3}{x}} \]
Push the divisor inside the root as a square
Why: Since the input is positive, it equals the square root of its own square, so it can move inside without an absolute value.
\[ \frac{\sqrt{4x^2+1}}{x} = \frac{\sqrt{4x^2+1}}{\sqrt{x^2}} = \sqrt{4 + \frac{1}{x^2}} \]
Take the limit of each piece
Why: The reciprocal square dies inside the root, and the three over the input dies in the denominator.
\[ = \frac{\sqrt{4 + 0}}{1 + 0} = \frac{2}{1} = 2 \]
Verify with a table, and notice the approach is slow
Why: The plus three in the denominator matters for a long time, so the values crawl toward two - which is exactly what an asymptote is allowed to do.
| input | value of the fraction |
|---|---|
| 100 | 1.94177 |
| 1000 | 1.99402 |
| 1000000 | 1.999994 |
Picture it
Animation
Shows: Each line of the worked example "a radical at the right end", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The plus three in the denominator matters for a long time, so the values crawl toward two - which is exactly what an asymptote is allowed to do.
Ranking
Put in order
Put the moves of Worked example: the same function at the left end 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. Now the input is not equal to the square root of its square, so the absolute value will produce a minus sign.
Worked example
Same function, other end. Only one step changes, and it changes the answer completely.
\[ \lim_{x \to -\infty} \frac{\sqrt{4x^2 + 1}}{x + 3} \]
Record the direction: the input is large and negative
Why: Now the input is not equal to the square root of its square, so the absolute value will produce a minus sign.
Divide top and bottom by the input, and move it inside the root correctly
Why: The root of the square is the absolute value, and for a negative input the absolute value is the opposite of the input.
\[ \frac{\sqrt{4x^2+1}}{x} = \frac{|x|\sqrt{4 + \frac{1}{x^2}}}{x} = \frac{-x\sqrt{4+\frac{1}{x^2}}}{x} = -\sqrt{4 + \frac{1}{x^2}} \]
The denominator behaves as before
Why: Dividing the denominator by the input gives one plus three over the input, which still tends to one; no absolute value is involved there.
\[ \frac{x+3}{x} = 1 + \frac{3}{x} \longrightarrow 1 \]
Assemble the limit
Why: The minus sign from the absolute value survives the whole computation and lands on the answer.
\[ \lim_{x \to -\infty} \frac{\sqrt{4x^2+1}}{x+3} = -2 \]
Verify numerically that the values really are negative
Why: A positive square root divided by a negative denominator must be negative, and the table agrees with the algebra rather than with the shortcut.
| input | value of the fraction |
|---|---|
| -100 | -2.06188 |
| -1000 | -2.00602 |
| -1000000 | -2.000006 |
Picture it
Animation
Shows: Each line of the worked example "the same function at the left end", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A positive square root divided by a negative denominator must be negative, and the table agrees with the algebra rather than with the shortcut.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Pull the input straight out of the root
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Treating the root of the square as the input itself is only legal when the input is nonnegative, and here it is very negative.
Treating the root of the square as the input itself is only legal when the input is nonnegative, and here it is very negative.
Why: Treating the root of the square as the input itself is only legal when the input is nonnegative, and here it is very negative.
Trap
\[ \lim_{x \to -\infty} \frac{\sqrt{9x^2 + 1}}{4x - 1} \]
Pull the input straight out of the root
Why: Treating the root of the square as the input itself is only legal when the input is nonnegative, and here it is very negative.
\[ \sqrt{9x^2+1} = x\sqrt{9 + \tfrac{1}{x^2}} \quad \text{(wrong)} \]
Report a positive answer
Why: The sign error propagates untouched to the final line - and a positive quantity divided by a negative one cannot be positive.
\[ \frac{3}{4} \quad \text{(wrong)} \]
\[ \lim_{x \to -\infty} \frac{\sqrt{9x^2 + 1}}{4x - 1} \]
Pull out the absolute value, then resolve it using the direction
Why: The input is negative, so its absolute value is the opposite of the input, and a minus sign appears.
\[ \sqrt{9x^2+1} = |x|\sqrt{9+\tfrac{1}{x^2}} = -x\sqrt{9+\tfrac{1}{x^2}} \]
Divide through and finish
Why: The numerator over the input becomes the negative of the root; the denominator over the input tends to four.
\[ = \lim_{x \to -\infty} \frac{-\sqrt{9 + \tfrac{1}{x^2}}}{4 - \tfrac{1}{x}} = -\frac{3}{4} \]
| input | value of the fraction |
|---|---|
| -1000 | -0.74981 |
| -1000000 | -0.7499998 |
Comparison
Comparison matrix
From Trap: losing the absolute value at the left end: refill the value of the fraction column from what you know. The rest of the table is as it appeared.
| input | value of the fraction |
|---|---|
| -1000 | -0.74981 |
| -1000000 | -0.7499998 |
Check
Write down the sign of the input first, then divide through.
\[ \lim_{x \to -\infty} \frac{\sqrt{9x^2 + 4}}{2x - 5} \]
Check your understanding
Evaluate the limit of sqrt(9x^2 + 4)/(2x - 5) as x runs to negative infinity.
Answer: A
Why: The root of 9x^2 is the absolute value of 3x, which equals -3x for negative inputs, so dividing through by x gives -3 over 2. Spot-check at x = -1000: the value is about 3000.0007/(-2005), which is about -1.4963.
Concept
When two pieces both grow without bound and you subtract them, the difference can be anything: zero, a number, or another runaway.
So a difference of two growing radicals has to be rewritten before it can be evaluated. The tool is the conjugate.
\[ \left(\sqrt{A} - B\right)\cdot\frac{\sqrt{A}+B}{\sqrt{A}+B} = \frac{A - B^2}{\sqrt{A}+B} \]
Multiplying by the conjugate over itself is multiplying by one, so nothing changes value. What changes is the shape: the subtraction moves upstairs, where the big terms cancel, and the sum downstairs is safely a sum.
Picture it
Animation
Shows: A runaway minus a runaway tells you nothing — a rendered Manim animation.
Rendered with Manim.
Takeaway: The form is a question. The algebra is the answer.
Estimation
Predict first
Both pieces run away, so there is nothing to substitute and no fraction to divide yet.
Commit before you compute: what does Worked example: a difference of runaways 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 large 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. The computed differences close in on one half, so the conjugate move did not change the value it only changed the form.
Worked example
\[ \lim_{x \to \infty} \left( \sqrt{x^2 + x} - x \right) \]
Both pieces run away, so there is nothing to substitute and no fraction to divide yet.
Multiply by the conjugate over itself
Why: This is multiplication by one, so the value is untouched; the point is to force the squares to cancel upstairs.
\[ = \lim_{x \to \infty} \frac{\left(\sqrt{x^2+x}-x\right)\left(\sqrt{x^2+x}+x\right)}{\sqrt{x^2+x}+x} \]
Expand the numerator as a difference of squares
Why: The root and the square undo each other, and the squared terms cancel, leaving a single linear term.
\[ = \lim_{x \to \infty} \frac{(x^2+x) - x^2}{\sqrt{x^2+x}+x} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2+x}+x} \]
Now it is an ordinary end-behavior fraction, so divide by the input
Why: The input is positive here, so it moves inside the root as its own square with no absolute value needed.
\[ = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \tfrac{1}{x}} + 1} \]
Send the reciprocal to zero
Why: The root tends to one, so the denominator tends to two.
\[ = \frac{1}{1+1} = \frac{1}{2} \]
Verify with large inputs
Why: The computed differences close in on one half, so the conjugate move did not change the value it only changed the form.
| input | value of the difference |
|---|---|
| 100 | 0.498756 |
| 1000 | 0.499875 |
| 10000 | 0.4999875 |
Picture it
Animation
Shows: Each line of the worked example "a difference of runaways", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The computed differences close in on one half, so the conjugate move did not change the value it only changed the form.
Check
Multiply by the conjugate over itself, cancel the squared terms, then divide through.
\[ \lim_{x \to \infty} \left( \sqrt{x^2 + 6x} - x \right) \]
Check your understanding
Evaluate the limit of sqrt(x^2 + 6x) - x as x grows without bound.
Answer: A
Why: The conjugate turns the difference into 6x over the sum of sqrt(x^2 + 6x) and x, which after dividing by x becomes 6 over 2, that is 3. Spot-check at x = 1000: sqrt(1006000) - 1000 is about 2.9955.
Section
Section 5
Concept
A top-heavy rational function has no horizontal asymptote - but that does not mean it has no asymptote at all.
slant (oblique) asymptote — A non-horizontal line that the graph approaches at the far ends. A rational function has one exactly when the numerator's degree is one more than the denominator's.
You find it with polynomial long division: divide the bottom into the top, and the quotient is the line.
\[ \frac{N(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)} \]
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of limit at infinity, horizontal asymptote, the root-of-a-square identity, slant (oblique) asymptote as Limits at Infinity, End Behavior, and Asymptotes uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Picture it
Animation
Shows: Top-heavy by one degree gives a slanted line — a rendered Manim animation.
Rendered with Manim.
Takeaway: Long division splits it into a line plus a remainder that fades.
Intuition
After the division you have a polynomial plus a leftover fraction. The leftover has a smaller degree on top than on the bottom, so it is bottom-heavy - it dies at both ends.
So far out, the function is the quotient plus almost nothing. The graph and the quotient become indistinguishable.
\[ f(x) - Q(x) = \frac{R(x)}{D(x)} \longrightarrow 0 \]
That is the actual definition of an asymptote: the difference between the graph and the line shrinks to nothing. A horizontal asymptote is just the case where the quotient happens to be a constant.
Step zero
Discussion prompt
Worked example: finding a slant asymptote — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: First division step: how many times does the bottom's leading term go…
Answer:
Worked example
\[ f(x) = \frac{2x^2 + 3x - 1}{x + 1} \]
The top is degree two and the bottom is degree one, so the gap is exactly one and a slant asymptote exists.
First division step: how many times does the bottom's leading term go into the top's?
Why: The squared term divided by the first power gives twice the input, which becomes the first term of the quotient.
\[ 2x \cdot (x+1) = 2x^2 + 2x \]
Subtract and bring down
Why: Removing that product leaves a smaller remainder to keep dividing.
\[ (2x^2 + 3x - 1) - (2x^2 + 2x) = x - 1 \]
Second division step
Why: The first power divided by the first power gives one, and subtracting the product leaves a constant remainder that cannot be divided further.
\[ 1 \cdot (x+1) = x + 1, \qquad (x-1)-(x+1) = -2 \]
Write the function in quotient-plus-remainder form
Why: The remainder term is bottom-heavy, so it vanishes at both ends and leaves the line behind.
\[ f(x) = 2x + 1 - \frac{2}{x+1} \]
Name the asymptote
Why: The quotient is the line the graph hugs at both ends.
\[ y = 2x + 1 \]
Verify by multiplying back and by measuring the gap
Why: The division reconstructs the original numerator exactly, and the measured gaps match the remainder term to the digit.
\[ (2x+1)(x+1) - 2 = 2x^2 + 3x + 1 - 2 = 2x^2 + 3x - 1 \quad \checkmark \]
| input | value of f | value on the line | gap |
|---|---|---|---|
| 100 | 200.9802 | 201 | -0.0198 |
| 1000 | 2000.9980 | 2001 | -0.0020 |
| -100 | -198.9798 | -199 | 0.0202 |
Picture it
Animation
Shows: Each line of the worked example "finding a slant asymptote", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The division reconstructs the original numerator exactly, and the measured gaps match the remainder term to the digit.
Concept
Nothing breaks - the division still works, the quotient is just no longer a line.
\[ \frac{x^3 + 1}{x} = x^2 + \frac{1}{x} \]
The graph hugs a parabola at the ends instead of a line. Courses usually call that a curved or parabolic asymptote and only require the linear case.
The takeaway is the same either way: divide, then read the polynomial part. It is the end behavior.
Commit first
Predict first
What is the slant asymptote of f(x) = (x^2 + 4x + 5)/(x + 1)?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: y = x + 3
Why: Long division gives x + 3 with remainder 2, so f(x) = x + 3 + 2/(x + 1) and the remainder term dies at both ends. Multiplying back: (x + 3)(x + 1) + 2 = x^2 + 4x + 3 + 2 = x^2 + 4x + 5.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
Do the division rather than guessing from the coefficients.
\[ f(x) = \frac{x^2 + 4x + 5}{x + 1} \]
Check your understanding
What is the slant asymptote of f(x) = (x^2 + 4x + 5)/(x + 1)?
Answer: A
Why: Long division gives x + 3 with remainder 2, so f(x) = x + 3 + 2/(x + 1) and the remainder term dies at both ends. Multiplying back: (x + 3)(x + 1) + 2 = x^2 + 4x + 3 + 2 = x^2 + 4x + 5.
Section
Section 6
Intuition
Squaring feels fast. Doubling feels slow. For small inputs, squaring is indeed ahead.
But doubling multiplies by the same factor every single step, while squaring's growth factor shrinks toward one. Compounding always wins a long enough race.
| input | the input squared | two raised to the input |
|---|---|---|
| 1 | 1 | 2 |
| 5 | 25 | 32 |
| 10 | 100 | 1024 |
| 20 | 400 | 1048576 |
| 30 | 900 | 1073741824 |
By an input of thirty the exponential is over a million times larger. Raise the polynomial's power and the exponential still wins - it just takes longer.
Pattern
Step through it
Step through Doubling always beats squaring, eventually one row at a time. What is driving the change, and what would the row after the last one be?
Picture it
Animation
Shows: Doubling always beats squaring — a rendered Manim animation.
Rendered with Manim.
Takeaway: The polynomial leads early and loses permanently.
Concept
For end-behavior purposes there is a strict pecking order. Anything higher in the list eventually swamps anything lower, no matter the constants involved.
\[ \lim_{x \to \infty} \frac{x^{p}}{e^{x}} = 0 \quad \text{and} \quad \lim_{x \to \infty} \frac{\ln x}{x^{p}} = 0 \quad \text{for every } p>0 \]
You can use these as facts here. Later, with a rule for indeterminate forms, you will be able to prove them in a line.
Explain it
Discussion prompt
Explain The growth hierarchy 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:
For end-behavior purposes there is a strict pecking order. Anything higher in the list eventually swamps anything lower, no matter the constants involved.
Picture it
Animation
Shows: Logarithmic, linear, quadratic and exponential curves drawn together.
Watch where the ordering becomes permanent.
Takeaway: Past some input the ordering is permanent: exponential beats every power, and every power beats every logarithm, whatever the constants suggest early on.
Worked example
\[ \lim_{x \to \infty} \frac{3e^{x} + 5x^2}{e^{x} - x} \]
Identify the fastest-growing term anywhere in the fraction
Why: The exponential outranks both polynomial terms, so it plays the role the highest power played for rational functions.
Divide every term by that fastest term
Why: Same move as before: dividing top and bottom by the same nonzero quantity leaves the value alone and turns everything else into something that dies.
\[ = \lim_{x \to \infty} \frac{3 + \dfrac{5x^2}{e^{x}}}{1 - \dfrac{x}{e^{x}}} \]
Apply the hierarchy to each leftover
Why: A power over an exponential tends to zero, and that is true for the squared term and the linear term alike.
\[ \frac{5x^2}{e^x} \longrightarrow 0, \qquad \frac{x}{e^x} \longrightarrow 0 \]
Read the answer
Why: Only the coefficients on the exponential survive, so the limit is their ratio.
\[ = \frac{3 + 0}{1 - 0} = 3 \]
Verify numerically at two inputs
Why: By an input of twenty the value already agrees with three to six decimal places, which is the hierarchy doing exactly what it claims.
| input | value of the fraction |
|---|---|
| 10 | 3.024073 |
| 20 | 3.0000042 |
Picture it
Animation
Shows: Each line of the worked example "the exponential decides it", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: By an input of twenty the value already agrees with three to six decimal places, which is the hierarchy doing exactly what it claims.
Concept
The natural exponential is not symmetric. It explodes to the right and collapses to the left.
\[ \lim_{x \to \infty} e^{x} = \infty, \qquad \lim_{x \to -\infty} e^{x} = 0 \]
So the exponential curve has a horizontal asymptote at the x-axis on the left only. That one-sided behavior is what makes it the backbone of every growth and decay model.
Whenever the exponent carries a minus sign, the two ends swap roles - so always simplify the exponent before deciding which end explodes.
Analogy
Discussion prompt
Explain The exponential at the other end 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 natural exponential is not symmetric. It explodes to the right and collapses to the left.
Estimation
Predict first
This shape models a population that grows quickly and then levels off at a carrying capacity, and it is the output layer of most simple classifiers.
Commit before you compute: what does Worked example: the logistic curve's two ceilings 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 at both ends
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. The outputs are squeezed between zero and one and press against each ceiling, exactly as the two limits predict.
Worked example
This shape models a population that grows quickly and then levels off at a carrying capacity, and it is the output layer of most simple classifiers.
\[ L(x) = \frac{1}{1 + e^{-x}} \]
Right end: decide what the exponent does
Why: As the input grows, the exponent is a large negative number, so the exponential collapses to zero.
\[ x \to \infty \;\Longrightarrow\; e^{-x} \to 0 \]
Substitute that into the fraction
Why: With the exponential gone the denominator is just one, so the whole expression settles at one.
\[ \lim_{x \to \infty} L(x) = \frac{1}{1+0} = 1 \]
Left end: the exponent flips sign
Why: As the input runs to negative infinity the exponent is a large positive number, so the exponential explodes and the denominator does too.
\[ \lim_{x \to -\infty} L(x) = 0 \]
Report both asymptotes
Why: Two ends, two different finite limits, so this function has two horizontal asymptotes - and they bracket every possible output.
\[ y = 1 \;\text{(right end)}, \qquad y = 0 \;\text{(left end)} \]
Verify with values at both ends
Why: The outputs are squeezed between zero and one and press against each ceiling, exactly as the two limits predict.
| input | value of the logistic |
|---|---|
| -10 | 0.0000454 |
| -5 | 0.006693 |
| 0 | 0.5 |
| 5 | 0.993307 |
| 10 | 0.9999546 |
Picture it
Animation
Shows: Each line of the worked example "the logistic curve's two ceilings", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The outputs are squeezed between zero and one and press against each ceiling, exactly as the two limits predict.
Check
There are exponentials on both the top and the bottom here, and large powers trying to distract you.
\[ \lim_{x \to \infty} \frac{x^5 + e^{x}}{2e^{x} + x^{10}} \]
Check your understanding
Evaluate the limit of (x^5 + e^x)/(2e^x + x^10) as x grows without bound.
Answer: A
Why: Divide every term by the exponential. Both power-over-exponential terms tend to zero, leaving 1 over 2. Spot-check at x = 50: the value is about 0.49999, since the tenth power is under two hundredths of a percent of the exponential there.
Section
Section 7
Concept
Here the input approaches a fixed number and the output runs away. That is the mirror image of everything in this deck so far.
\[ \lim_{x \to a^{+}} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^{-}} f(x) = \pm\infty \]
One side blowing up is enough to make the line a vertical asymptote. The two sides may run in opposite directions, and often do.
Keeping the two ideas straight is easy if you read the arrow: an arrow pointing at a number asks about a vertical asymptote, an arrow pointing off the page asks about a horizontal one.
Counterexample
Discussion prompt
One side blowing up is enough to make the line a vertical asymptote. The two sides may run in opposite directions, and often do.
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:
Keeping the two ideas straight is easy if you read the arrow: an arrow pointing at a number asks about a vertical asymptote, an arrow pointing off the page asks about a horizontal one.
Picture it
Animation
Shows: A curve spiking near a vertical asymptote.
Check the numerator before calling it a wall.
Takeaway: A vertical asymptote needs the denominator to approach zero while the numerator does not. If both vanish, you have a hole to investigate instead.
Concept
A zero in the denominator makes the function undefined - but undefined comes in two flavors, and only one of them is a wall.
Factor the top and the bottom completely, then cancel. What happens to a factor tells you which flavor you have.
| The factor after cancelling | What is at that input | Why |
|---|---|---|
| still in the denominator | vertical asymptote | the bottom still goes to zero while the top does not |
| cancelled away completely | a hole | top and bottom both vanish at the same rate, so the value is finite |
A hole is an invisible single missing point. The graph looks perfectly ordinary there; the function simply has no output at that one input.
Comparison
Comparison matrix
From Which zeros of the denominator are real walls?: refill the What is at that input column from what you know. The rest of the table is as it appeared.
| The factor after cancelling | What is at that input | Why |
|---|---|---|
| still in the denominator | vertical asymptote | the bottom still goes to zero while the top does not |
| cancelled away completely | a hole | top and bottom both vanish at the same rate, so the value is finite |
Picture it
Animation
Shows: Which zeros of the denominator are real walls? — a rendered Manim animation.
Rendered with Manim.
Takeaway: Cancel first. Only the survivors are walls.
Step zero
Discussion prompt
Worked example: one hole, one wall — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Factor the top and the bottom completely
Answer:
Worked example
\[ f(x) = \frac{x^2 - 4}{x^2 - x - 2} \]
Find every vertical asymptote, every hole, and the horizontal asymptote.
Factor the top and the bottom completely
Why: Nothing about holes or walls can be decided before the shared factors are visible.
\[ f(x) = \frac{(x-2)(x+2)}{(x-2)(x+1)} \]
List the inputs that break the denominator
Why: Those are the only candidates. Every vertical asymptote and every hole must come from this list.
\[ x = 2 \quad \text{and} \quad x = -1 \]
Cancel the shared factor and record the restriction
Why: The cancelled form agrees with the original everywhere except at the cancelled input, where the original is still undefined.
\[ f(x) = \frac{x+2}{x+1}, \quad x \neq 2 \]
Classify each candidate
Why: The factor for the input two cancelled, so that is a hole; the factor for negative one survived, so that is a vertical asymptote.
\[ \text{hole at } \left(2, \tfrac{4}{3}\right), \qquad \text{vertical asymptote } x = -1 \]
Get the horizontal asymptote from the degrees
Why: Top and bottom are both degree two with leading coefficient one, so the ends level off at height one.
\[ y = 1 \]
Verify all three claims with nearby values
Why: Near the hole the values sit calmly beside four thirds; near the wall they explode with opposite signs on the two sides; far out they close in on one.
| input | value of f | what it confirms |
|---|---|---|
| 1.99 | 1.33445 | hole, value near 4/3 |
| -0.99 | 101.00 | wall, right side runs up |
| -1.01 | -99.00 | wall, left side runs down |
| 1000 | 1.000999 | horizontal asymptote at height 1 |
Picture it
Animation
Shows: Each line of the worked example "one hole, one wall", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Near the hole the values sit calmly beside four thirds; near the wall they explode with opposite signs on the two sides; far out they close in on one.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Set the denominator to zero and report both answers as vertical asymptotes
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The rule 'denominator zero means vertical asymptote' was applied without factoring, so the shared factor was never noticed.
The rule 'denominator zero means vertical asymptote' was applied without factoring, so the shared factor was never noticed.
Why: The rule 'denominator zero means vertical asymptote' was applied without factoring, so the shared factor was never noticed.
Trap
\[ f(x) = \frac{x-3}{x^2-9} \]
Set the denominator to zero and report both answers as vertical asymptotes
Why: The rule 'denominator zero means vertical asymptote' was applied without factoring, so the shared factor was never noticed.
\[ x^2 - 9 = 0 \;\Longrightarrow\; x = 3 \text{ and } x = -3 \quad \text{(both called walls: wrong)} \]
Predict a blow-up near the input three
Why: The prediction is checkable, and it fails: the actual values there are around one sixth, nowhere near infinite.
| input | predicted | actual value |
|---|---|---|
| 2.9 | huge | 0.169492 |
| 3.1 | huge | 0.163934 |
\[ f(x) = \frac{x-3}{x^2-9} = \frac{x-3}{(x-3)(x+3)} \]
Cancel the shared factor first, then classify
Why: The factor for the input three cancels completely, so that input is a hole, not a wall; only the surviving factor makes an asymptote.
\[ f(x) = \frac{1}{x+3}, \quad x \neq 3 \]
Report the hole with its height and the single wall
Why: The height of the hole comes from the cancelled form evaluated at the missing input.
\[ \text{hole at } \left(3, \tfrac{1}{6}\right), \qquad \text{vertical asymptote } x = -3 \]
Verify against real values on both candidates
Why: The values near three are calm and near one sixth; the values near negative three really do explode.
| input | value of f |
|---|---|
| 2.9 | 0.169492 |
| 3.1 | 0.163934 |
| -2.9 | 10.00 |
| -2.99 | 100.00 |
Pattern
Step through it
Step through Trap: every zero of the denominator is a wall one row at a time. What is driving the change, and what would the row after the last one be?
Ranking
Put in order
These are the steps of The complete asymptote checklist, 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 rational function, run these in order and you will have described the whole picture.
Edge cases
Discussion prompt
The complete asymptote checklist 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:
Given a rational function, run these in order and you will have described the whole picture.
Ranking
Put in order
Put the moves of Worked example: the full picture 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. Both quadratics factor, and they share a factor, so a hole is coming.
Worked example
\[ f(x) = \frac{x^2 - 9}{x^2 - 2x - 3} \]
Describe every hole, every asymptote, the intercepts, and whether the graph ever crosses its horizontal asymptote.
Factor and cancel
Why: Both quadratics factor, and they share a factor, so a hole is coming.
\[ f(x) = \frac{(x-3)(x+3)}{(x-3)(x+1)} = \frac{x+3}{x+1}, \quad x \neq 3 \]
Read off the hole
Why: The cancelled input is missing from the domain, and its height comes from the reduced form.
\[ \text{hole at } \left(3, \tfrac{3}{2}\right) \]
Read off the vertical asymptote
Why: The surviving denominator factor vanishes at negative one while the numerator does not, so the outputs run away there.
\[ x = -1 \]
Read off the horizontal asymptote
Why: In the reduced form both parts are degree one with leading coefficient one, so both ends level off at height one.
\[ y = 1 \]
Find the intercepts
Why: The graph meets the x-axis where the reduced numerator vanishes, and meets the y-axis at the output for input zero.
\[ x\text{-intercept: } (-3, 0), \qquad y\text{-intercept: } (0, 3) \]
Test for a crossing of the horizontal asymptote
Why: Setting the function equal to one produces a contradiction, so this graph approaches its asymptote without ever meeting it.
\[ \frac{x+3}{x+1} = 1 \;\Longrightarrow\; x+3 = x+1 \;\Longrightarrow\; 3 = 1 \;\text{(no solution)} \]
Verify every claim against the original function
Why: The intercept, the hole height, and the far-field value all check out against the unsimplified formula, so no step lost information.
| input | value of f | what it confirms |
|---|---|---|
| 0 | 3 | y-intercept |
| 2.99 | 1.501253 | hole height near 3/2 |
| -0.99 | 201.00 | vertical asymptote at negative one |
| 1000 | 1.001998 | horizontal asymptote at height 1 |
Picture it
Animation
Shows: Each line of the worked example "the full picture", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The intercept, the hole height, and the far-field value all check out against the unsimplified formula, so no step lost information.
Prediction
Predict first
Which description of f(x) = (x^2 - 1)/(x^2 - 3x + 2) is correct?
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: Hole at x = 1, vertical asymptote at x = 2, horizontal asymptote y = 1
Why: Factoring gives (x - 1)(x + 1) over (x - 1)(x - 2), which reduces to (x + 1)/(x - 2) with the input 1 removed. The cancelled factor makes a hole at height -2, the surviving factor makes the wall at 2, and equal degrees give the horizontal asymptote at height 1.
Check
Factor first, cancel second, classify third.
\[ f(x) = \frac{x^2 - 1}{x^2 - 3x + 2} \]
Check your understanding
Which description of f(x) = (x^2 - 1)/(x^2 - 3x + 2) is correct?
Answer: A
Why: Factoring gives (x - 1)(x + 1) over (x - 1)(x - 2), which reduces to (x + 1)/(x - 2) with the input 1 removed. The cancelled factor makes a hole at height -2, the surviving factor makes the wall at 2, and equal degrees give the horizontal asymptote at height 1.
Elimination
Eliminate the wrong options
At which input does the graph of f(x) = (4x^2 + 8x)/(x^2 + 4) cross its horizontal asymptote y = 4?
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: Setting the fraction equal to 4 gives 4x^2 + 8x = 4x^2 + 16, so 8x = 16 and the input is 2. Substituting back: (16 + 16)/(4 + 4) = 32/8 = 4, so the point really is on the line.
Check
The horizontal asymptote of this function is the line at height four. Solve, do not guess.
\[ f(x) = \frac{4x^2 + 8x}{x^2 + 4} \]
Check your understanding
At which input does the graph of f(x) = (4x^2 + 8x)/(x^2 + 4) cross its horizontal asymptote y = 4?
Answer: A
Why: Setting the fraction equal to 4 gives 4x^2 + 8x = 4x^2 + 16, so 8x = 16 and the input is 2. Substituting back: (16 + 16)/(4 + 4) = 32/8 = 4, so the point really is on the line.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — What "At Infinity" Actually Means · Rational Functions: The Highest Power Wins · Horizontal Asymptotes, Up Close · Radicals and the Sign Trap · When the Asymptote Tilts · Growth Rates: Who Wins the Race. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
End behavior is one question asked at two ends, and one technique answers almost all of it: divide out the fastest-growing thing and see what survives.
| Question | First move | Answer looks like |
|---|---|---|
| end behavior of a rational function | divide by the denominator's highest power | a number, or growth with a sign |
| radical at the left end | write the root of the square as an absolute value | a number, usually negative |
| difference of two radicals | multiply by the conjugate over itself | a finite number |
| top-heavy by exactly one degree | polynomial long division | a slanted line |
| wall or hole at a bad input | factor and cancel | a hole with a height, or a vertical asymptote |
Next up: the derivative itself - the limit that turns a curve's steepness into a number. Everything you just practiced about limits is the machinery it runs on.
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