Evaluate each limit and state what it says about the graph. (a) $\displaystyle\lim_{x\to-\infty}\frac{7x-4}{x^{2}+3}$ (b) $\displaystyle\lim_{x\to\infty}\frac{2x^{3}-x}{5x^{2}+4}$

(a) $0$; horizontal asymptote $y=0$. (b) $+\infty$; no horizontal asymptote — the graph follows the slant asymptote $y=\frac25x$.

For $f(x)=\dfrac{2x+1}{x-3}$, find every vertical and horizontal asymptote, and compute the one-sided limits at the vertical asymptote.

Vertical asymptote $x=3$: $f\to+\infty$ as $x\to3^{+}$ and $f\to-\infty$ as $x\to3^{-}$. Horizontal asymptote $y=2$ in both directions.

Find all horizontal asymptotes of $f(x)=\dfrac{\sqrt{9x^{2}+2x}}{4x+3}$.

Two: $y=\frac34$ as $x\to\infty$ and $y=-\frac34$ as $x\to-\infty$.

Evaluate $\displaystyle\lim_{x\to\infty}\frac{2x+\sin x}{x+1}$, and explain why l'Hôpital's Rule cannot finish this problem.

The limit is $2$. l'Hôpital gives $\frac{2+\cos x}{1}$, which has no limit, so the rule reaches no conclusion.

Find the slant asymptote and the vertical asymptote of $f(x)=\dfrac{2x^{2}-3x+1}{x-2}$, and say on which side of the slant asymptote the curve lies as $x\to+\infty$ and as $x\to-\infty$.

Slant asymptote $y=2x+1$; vertical asymptote $x=2$ with $f\to+\infty$ from the right and $f\to-\infty$ from the left. The curve is above the line as $x\to+\infty$ and below it as $x\to-\infty$.

Find the constant $a$ for which $\displaystyle\lim_{x\to\infty}\left(\sqrt{x^{2}+ax}-\sqrt{x^{2}+1}\right)=3$. Why is the argument "both roots behave like $x$, so the limit is $0$" wrong?

$a=6$. The roots are $x+\frac{a}{2}+(\text{small})$ and $x+(\text{small})$: the $x$ terms cancel but the constants do not.

Limits & Continuity · Topic 04 of 23

Limits at Infinity and Asymptotes

x

5 min readCore — on every final8 formulas6 quick checks

Key ideas

5 things to remember

1
$Divide by the highest power downstairs$ $x$
2
$\sqrt{x^2}=|x|$ $x$
3
$\infty-\infty$ $\sqrt{x^2+5x}-x$
4
$Vertical asymptotes: cancel, then check signs$ $0$
5
$Slant asymptotes come from long division$ $f(x)=mx+b+\frac{r(x)}{q(x)}$

f(x)=\frac{2x+1}{x-1}=2+\frac{3}{x-1}

f(x)=\frac{3x+2}{\sqrt{2x^2+1}}

Formulas

What to have memorised

$The basic limit at infinity$
$\lim_{x\to\pm\infty}\frac{c}{x^{p}}=0\qquad(p>0)$
$x$
$n$
$\lim_{x\to\pm\infty}\frac{a_nx^{n}+\cdots}{b_dx^{d}+\cdots}=\begin{cases}0 & n<d\\ \dfrac{a_n}{b_d} & n=d\\ \pm\infty & n>d\end{cases}$
$x^d$
$Square root of a square$
$\sqrt{x^{2}}=|x|=\begin{cases}x, & x\ge 0\\ -x, & x<0\end{cases}$
$\sqrt{ax^2+bx+c}=|x|\sqrt{a+\frac{b}{x}+\frac{c}{x^2}}$
$\infty-\infty$
$\sqrt{A}-\sqrt{B}=\frac{A-B}{\sqrt{A}+\sqrt{B}}$
$x$
$x=a$
$\lim_{x\to a^{\pm}}\frac{p(x)}{q(x)}=\pm\infty\quad\text{if } q(a)=0,\ p(a)\neq 0$
$a$
$y=mx+b$
$m=\lim_{x\to\infty}\frac{f(x)}{x},\qquad b=\lim_{x\to\infty}\bigl(f(x)-mx\bigr)$
$n=d+1$
$Bounded over unbounded$
$-\frac{1}{x}\le\frac{\sin x}{x}\le\frac{1}{x}\ (x>0)\ \Longrightarrow\ \lim_{x\to\infty}\frac{\sin x}{x}=0$
$\to\infty$
$x\to\infty$
$(\ln x)^{q}\ll x^{p}\ll e^{cx}\qquad(p,\ c>0)$
$\to0$

Find every asymptote of a function

The steps, in order

1 $Factor numerator and denominator completely and cancel common factors. A zero of the denominator that cancels completely is a hole, not an asymptote.$
2 $a$
3 $n<d$
4 $n=d+1$
5 $\sqrt{x^{2}}=|x|$
6 $y=2$

Watch out

The mistakes that cost marks

✗ Wrong
$\sqrt{9x^2+1}\approx 3x$
✓ Right
$x<0$
$\sqrt{x^2}=|x|$
✗ Wrong
$\lim_{x\to\infty}\left(\sqrt{x^2+5x}-x\right)=\infty-\infty=0$
✓ Right
$\frac{5x}{\sqrt{x^2+5x}+x}=\frac{5}{\sqrt{1+5/x}+1}\to\frac52$
$\infty-\infty$
✗ Wrong
$\frac{x+2}{x^2-4}$
✓ Right
$\frac{x+2}{(x-2)(x+2)}=\frac{1}{x-2}$
✗ Wrong
$\lim_{x\to2}\frac{1}{x-2}=\infty$
✓ Right
$\lim_{x\to2^{+}}\frac{1}{x-2}=+\infty$
$Why: Check the sign of the denominator on each side.$
✗ Wrong
$\frac{2x+\sin x}{x+1}\to\frac{2+\cos x}{1}$
✓ Right
$x$
$\lim f'/g'$
✗ Wrong
$y=L$
✓ Right
$\frac{\sin x}{x}$
$Why: An asymptote only controls the far ends of the graph.$

Quick check

Commit to an answer before you reveal one

Q1easy
$\displaystyle\lim_{x\to-\infty}\frac{7x-4}{x^{2}+3}$
Hint
$x^2$
Show answer
$0$ $x^2$
Q2easy
$f(x)=\dfrac{2x+1}{x-3}$
Hint
$x-3$
Show answer
$x=3$ $x=3$
Q3medium
$f(x)=\dfrac{\sqrt{9x^{2}+2x}}{4x+3}$
Hint
$\sqrt{9x^2+2x}=|x|\sqrt{9+\frac{2}{x}}$
Show answer
$y=\frac34$ $x\neq0$
Q4medium
$\displaystyle\lim_{x\to\infty}\frac{2x+\sin x}{x+1}$
Hint
$x$
Show answer
$2$ $x>0$
Q5medium
$f(x)=\dfrac{2x^{2}-3x+1}{x-2}$
Hint
$Numerator degree is one more than denominator degree: long divide and look at the remainder term.$
Show answer
$y=2x+1$ $2x^{2}-3x+1=(2x+1)(x-2)+3$
Q6hard
$a$
Hint
$\infty-\infty$
Show answer
$a=6$ $\sqrt{x^{2}+ax}-\sqrt{x^{2}+1}=\frac{(x^{2}+ax)-(x^{2}+1)}{\sqrt{x^{2}+ax}+\sqrt{x^{2}+1}}=\frac{ax-1}{\sqrt{x^{2}+ax}+\sqrt{x^{2}+1}}.$

On the exam

How this topic is marked

\lim_{x\to a^{+}}

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Limits at Infinity and Asymptotes

Key ideas

Divide by the highest power downstairs

Under a root, $\sqrt{x^2}=|x|$

$\infty-\infty$ needs a conjugate

Vertical asymptotes: cancel, then check signs

Slant asymptotes come from long division

Formulas

Find every asymptote of a function

Watch out

Quick check

On the exam

Practise more

Read next

Stuck on this topic?

Navigation

Key ideas

Divide by the highest power downstairs

Under a root, x2=∣x∣\sqrt{x^2}=|x|x2​=∣x∣

∞−∞\infty-\infty∞−∞ needs a conjugate

Vertical asymptotes: cancel, then check signs

Slant asymptotes come from long division

Formulas

Find every asymptote of a function

Watch out

Quick check

On the exam

Practise more

Read next

Stuck on this topic?

Under a root, $\sqrt{x^2}=|x|$

$\infty-\infty$ needs a conjugate