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Calculus BC · Unit 1 · 1.15
AP Calculus · Limits and Continuity
Limits at Infinity and Horizontal Asymptotes
End behaviour is a race between numerator and denominator — and you must run it in both directions.
key idea$\lim_{x\to\pm\infty}f(x)=L\iff y=L\text{ is a horizontal asymptote}$
Reading the graph
A limit at infinity asks what happens to the output as $x$ runs off the end of the axis. If it settles on a number $L$, the line $y=L$ is a horizontal asymptote and describes the function's end behaviour.
degrees equal → $y=\tfrac32$
| $x$ | $10$ | $100$ | $1000$ | $\to\infty$ |
|---|
| $\dfrac{3x}{x^{2}+1}$ | $0.297$ | $0.0300$ | $0.00300$ | $\to 0$ |
| $\dfrac{3x^{2}+1}{2x^{2}+5}$ | $1.4926$ | $1.4999$ | $1.5000$ | $\to 1.5$ |
| $\dfrac{x^{2}}{x+4}$ | $7.14$ | $96.2$ | $996$ | $\to\infty$ |
The same three cases as numbers. Watch how fast each column settles — or fails to.
| Situation | $\lim_{x\to\infty}$ | Horizontal asymptote |
|---|
| denominator has higher degree | $0$ | $y=0$ |
| degrees equal, leading coefficients $a$ and $b$ | $\dfrac{a}{b}$ | $y=\dfrac{a}{b}$ |
| numerator has higher degree | $\pm\infty$ | none |
| $e^{-x}$ as $x\to\infty$ | $0$ | $y=0$ |
| $\arctan x$ as $x\to\infty$ | $\dfrac{\pi}{2}$ | $y=\dfrac{\pi}{2}$ |
| $\sin x$ as $x\to\infty$ | DNE — oscillates | none |
A function may have two different horizontal asymptotes — one as $x\to+\infty$ and another as $x\to-\infty$. Always check both directions.
Worked examples
EX 1
Divide by the highest power
$\dfrac{3x^{2}+1}{2x^{2}+5}$Find $\lim\limits_{x\to\infty}\dfrac{3x^{2}+1}{2x^{2}+5}$ and any horizontal asymptote.
1
Identify the highest power in the denominator
It is $x^{2}$.
2
Divide every term by $x^{2}$
$\dfrac{3+\frac{1}{x^{2}}}{2+\frac{5}{x^{2}}}$.
3
Send $x\to\infty$
Each $\dfrac{k}{x^{2}}\to0$, leaving $\dfrac{3+0}{2+0}$.
4
State the result
The limit is $\dfrac32$, so $y=\dfrac32$ is a horizontal asymptote.
5
Shortcut
Equal degrees means the answer is always the ratio of the leading coefficients — but show the division if a justification is requested.
EX 2
The denominator wins
$\dfrac{3x}{x^{2}+1}$Find $\lim\limits_{x\to\infty}\dfrac{3x}{x^{2}+1}$.
1
Compare degrees
Numerator degree $1$, denominator degree $2$. The bottom grows faster.
2
Divide by $x^{2}$
$\dfrac{\frac{3}{x}}{1+\frac{1}{x^{2}}}$.
3
Take the limit
$\dfrac{0}{1+0}=0$.
4
Conclude
$y=0$ is a horizontal asymptote. The curve rises, turns and decays back to the axis — and it crosses $y=0$ on the way, which is allowed.
EX 3
Two different asymptotes
$\dfrac{\sqrt{4x^{2}+1}}{x}$Find both end-behaviour limits of $h(x)=\dfrac{\sqrt{4x^{2}+1}}{x}$.
1
As $x\to+\infty$
For positive $x$, $\sqrt{x^{2}}=x$. Divide inside and out: $\dfrac{x\sqrt{4+\frac{1}{x^{2}}}}{x}\to 2$.
2
As $x\to-\infty$
For negative $x$, $\sqrt{x^{2}}=\lvert x\rvert=-x$. So the expression becomes $\dfrac{-x\sqrt{4+\frac{1}{x^{2}}}}{x}\to -2$.
3
State both
$y=2$ as $x\to+\infty$ and $y=-2$ as $x\to-\infty$ — two horizontal asymptotes.
4
The trap
Writing $\sqrt{x^{2}}=x$ for negative $x$ loses the second asymptote entirely. Roots of even powers carry an absolute value.
EX 4
No asymptote
$\dfrac{x^{2}}{x+4}$Does $\dfrac{x^{2}}{x+4}$ have a horizontal asymptote?
1
Compare degrees
Numerator degree $2$ exceeds denominator degree $1$.
2
Divide by $x$
$\dfrac{x}{1+\frac{4}{x}}$.
3
Take the limit
The numerator grows without bound while the denominator tends to $1$, so the quotient $\to+\infty$.
4
Conclude
No horizontal asymptote. The limit does not exist as a finite number, so no horizontal line describes the end behaviour.
Watch out
A graph may cross its horizontal asymptote
Unlike a vertical asymptote, there is no rule against it. $\dfrac{3x}{x^{2}+1}$ crosses $y=0$ at the origin and still has that asymptote.
Check both directions
$x\to+\infty$ and $x\to-\infty$ can give different values. Any expression with $\sqrt{x^{2}}$, $\lvert x\rvert$ or $e^{x}$ is a candidate.
$\sqrt{x^{2}}=\lvert x\rvert$, not $x$
For negative $x$ that sign flip is the whole answer — it is what produces the second asymptote in Example 3.
Oscillation is not a limit
$\sin x$ stays bounded between $-1$ and $1$ but never settles, so $\lim_{x\to\infty}\sin x$ does not exist and there is no asymptote.