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Concept Review

Differential Calculus

Limits at Infinity and Asymptotes

Limits at infinity describe the end behaviour of a graph: what f(x)f(x) does as xx runs off to the far right or the far left. Infinite limits describe the opposite phenomenon, blow-up at a finite point. Together they produce the horizontal, vertical and slant asymptotes that are the first things you draw when sketching a curve, and they are the language used later for improper integrals, series and growth comparisons. Almost every computation here reduces to two moves: spot the dominant term, then make the algebra legitimate by dividing by the right power of xx.

1. Limits at infinity: definition and basic facts

Definition. Let ff be defined on some interval (a,∞)(a,\infty). We write lim⁡x→∞f(x)=L\lim_{x\to\infty}f(x)=L if for every ε>0\varepsilon > 0 there is a number MM such that x>M⇒∣f(x)−L∣<εx > M \Rightarrow |f(x)-L| < \varepsilon. Similarly lim⁡x→−∞f(x)=L\lim_{x\to-\infty}f(x)=L requires ff defined on some (−∞,a)(-\infty,a) and x<M⇒∣f(x)−L∣<εx < M \Rightarrow |f(x)-L| < \varepsilon.

Definition (horizontal asymptote). The line y=Ly=L is a horizontal asymptote of y=f(x)y=f(x) if lim⁡x→∞f(x)=L\lim_{x\to\infty}f(x)=L or lim⁡x→−∞f(x)=L\lim_{x\to-\infty}f(x)=L. A function therefore has at most two horizontal asymptotes, one per direction. A graph is allowed to cross its horizontal asymptote, even infinitely often: f(x)=sin⁡xxf(x)=\dfrac{\sin x}{x} crosses y=0y=0 at x=kπx=k\pi for every nonzero integer kk.

Definition (infinite limit at infinity). lim⁡x→∞f(x)=∞\lim_{x\to\infty}f(x)=\infty means: for every NN there is MM with x>M⇒f(x)>Nx > M \Rightarrow f(x) > N. This is not a limit that exists; it is a precise description of how it fails.

Core fact. For every real p>0p > 0,   lim⁡x→∞1xp=0\;\lim_{x\to\infty}\dfrac{1}{x^{p}}=0. The same holds as x→−∞x\to-\infty whenever xpx^{p} is defined for negative xx (safe for pp a positive integer).

The limit laws (sum, difference, product, quotient with nonzero denominator limit) hold verbatim for x→±∞x\to\pm\infty provided each individual limit exists as a finite number. They say nothing about the indeterminate forms ∞−∞\infty-\infty, ∞∞\dfrac{\infty}{\infty}, 0⋅∞0\cdot\infty, 00\dfrac{0}{0}, which must be resolved algebraically first.

2. Infinite limits and vertical asymptotes

Definition. lim⁡x→af(x)=∞\lim_{x\to a}f(x)=\infty means: for every NN there is δ>0\delta > 0 with 0<∣x−a∣<δ⇒f(x)>N0 < |x-a| < \delta \Rightarrow f(x) > N. Replace the last inequality by f(x)<−Nf(x) < -N for −∞-\infty; restrict to a<x<a+δa < x < a+\delta or a−δ<x<aa-\delta < x < a for one-sided versions.

Definition (vertical asymptote). x=ax=a is a vertical asymptote of ff if at least one of lim⁡x→a+f(x)\lim_{x\to a^{+}}f(x), lim⁡x→a−f(x)\lim_{x\to a^{-}}f(x) equals +∞+\infty or −∞-\infty.

Recipe for a rational function f=pqf=\dfrac{p}{q}: factor and cancel all common factors first.

  • If, after cancelling, q(a)=0q(a)=0 and p(a)≠0p(a)\ne 0, then x=ax=a is a vertical asymptote. The form is nonzero0\dfrac{\text{nonzero}}{0}, so the magnitude is infinite; get the sign from a sign table just to the left and just to the right of aa.
  • If every copy of (x−a)(x-a) in the denominator cancels — that is, the numerator contains at least as many factors of (x−a)(x-a) as the denominator does — the point is a hole (removable discontinuity), not an asymptote, and the limit there is finite. Partial cancellation is not enough: in x−a(x−a)2=1x−a\dfrac{x-a}{(x-a)^{2}}=\dfrac{1}{x-a} a factor of (x−a)(x-a) does cancel, yet x=ax=a is still a vertical asymptote.
  • The form 00\dfrac{0}{0} never tells you the answer by itself; it only tells you to simplify.

3. End behaviour of rational functions

Write f(x)=anxn+⋯+a0bdxd+⋯+b0f(x)=\dfrac{a_nx^{n}+\cdots+a_0}{b_dx^{d}+\cdots+b_0} with an≠0a_n\ne 0, bd≠0b_d\ne 0. The dominant-term heuristic says f(x)≈anxnbdxdf(x)\approx \dfrac{a_nx^{n}}{b_dx^{d}} for large ∣x∣|x|; the rigorous version is to divide numerator and denominator by xdx^{d}, the highest power in the denominator, and then apply the limit laws. The quotient law actually finishes the job only when n≤dn\le d, so that the new numerator has a finite limit; when n>dn > d the new numerator is unbounded, and you argue instead that a quantity tending to ±∞\pm\infty divided by one tending to a nonzero finite limit tends to ±∞\pm\infty.

Caselim⁡x→±∞f(x)\lim_{x\to\pm\infty}f(x)Asymptote
n<dn < d00horizontal y=0y=0
n=dn = danbd\dfrac{a_n}{b_d}horizontal y=anbdy=\dfrac{a_n}{b_d}
n=d+1n = d+1+∞+\infty or −∞-\inftyslant y=mx+by=mx+b from long division
n>d+1n > d+1+∞+\infty or −∞-\inftyno line; a polynomial asymptote of degree n−dn-d

Whether an infinite limit is +∞+\infty or −∞-\infty depends on the sign of anbd\dfrac{a_n}{b_d} and, as x→−∞x\to-\infty, on the parity of n−dn-d.

4. Square roots: x2=∣x∣\sqrt{x^{2}}=|x|, not xx

The single most common error in this topic. Always use x2=∣x∣={x,x≥0−x,x<0\sqrt{x^{2}}=|x|=\begin{cases} x, & x\ge 0\\ -x, & x < 0\end{cases} so that for x<0x < 0, pulling xx out of a root flips the sign: ax2+bx+c=∣x∣a+bx+cx2=−xa+bx+cx2\sqrt{ax^{2}+bx+c}=|x|\sqrt{a+\dfrac{b}{x}+\dfrac{c}{x^{2}}}=-x\sqrt{a+\dfrac{b}{x}+\dfrac{c}{x^{2}}}.

Worked Example 1. Find both horizontal asymptotes of f(x)=3x+22x2+1f(x)=\dfrac{3x+2}{\sqrt{2x^{2}+1}}.

As x→+∞x\to+\infty divide top and bottom by xx, writing x=x2x=\sqrt{x^{2}} downstairs (legal since x>0x > 0):

\frac{3x+2}{\sqrt{2x^{2}+1}} &= \frac{\dfrac{3x+2}{x}}{\dfrac{\sqrt{2x^{2}+1}}{\sqrt{x^{2}}}} = \frac{3+\dfrac{2}{x}}{\sqrt{2+\dfrac{1}{x^{2}}}} \longrightarrow \frac{3+0}{\sqrt{2+0}}=\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}. \end{aligned}$$ As $x\to-\infty$ we have $x < 0$, so $x=-\sqrt{x^{2}}$ and dividing by $x$ introduces a minus sign inside the root: $$\begin{aligned} \frac{3x+2}{\sqrt{2x^{2}+1}} &= \frac{3+\dfrac{2}{x}}{-\sqrt{2+\dfrac{1}{x^{2}}}} \longrightarrow \frac{3}{-\sqrt{2}}=-\frac{3\sqrt{2}}{2}. \end{aligned}$$ So there are two horizontal asymptotes: $y=\dfrac{3\sqrt{2}}{2}$ and $y=-\dfrac{3\sqrt{2}}{2}$. ### 5. Slant (oblique) asymptotes **Definition.** The line $y=mx+b$ is an asymptote of $f$ as $x\to\infty$ if $\lim_{x\to\infty}\left[f(x)-(mx+b)\right]=0$. Equivalently, compute $m=\lim_{x\to\infty}\dfrac{f(x)}{x}$ and then $b=\lim_{x\to\infty}\left[f(x)-mx\right]$; the line exists exactly when both limits exist and are finite. The same definition applies as $x\to-\infty$, possibly with a **different** line. For a rational function with $n=d+1$, polynomial long division gives $f(x)=(mx+b)+\dfrac{r(x)}{q(x)}$ with $\text{deg }r < \text{deg }q$, so the remainder term tends to $0$ and $y=mx+b$ is the slant asymptote. The **sign of that remainder**, $\dfrac{r(x)}{q(x)}=f(x)-(mx+b)$, tells you whether the curve sits above the line (positive) or below it (negative). **Worked Example 2.** Analyse $f(x)=\dfrac{x^{2}+1}{x-1}$. Long division: $x^{2}+1=(x+1)(x-1)+2$, so $$f(x)=x+1+\frac{2}{x-1}.$$ Since $\dfrac{2}{x-1}\to 0$ as $x\to\pm\infty$, the line $y=x+1$ is a slant asymptote in both directions; the curve lies **above** it for $x > 1$ and **below** it for $x < 1$. The numerator at $x=1$ is $2\ne 0$, so $x=1$ is a vertical asymptote, with $\lim_{x\to 1^{+}}f(x)=+\infty$ and $\lim_{x\to 1^{-}}f(x)=-\infty$. There is no horizontal asymptote (a function cannot have a horizontal and a slant asymptote in the same direction). ### 6. Growth-rate hierarchy | Slowest to fastest | Representative | Fact (all limits as $x\to\infty$) | | --- | --- | --- | | logarithms | $(\ln x)^{q}$ | $\lim \dfrac{(\ln x)^{q}}{x^{p}}=0$ for all $p > 0$, all $q$ | | powers and roots | $x^{p}$ | $\lim \dfrac{x^{p}}{e^{cx}}=0$ for all $p > 0$, $c > 0$ | | exponentials | $a^{x}$ with $a > 1$ | $\lim \dfrac{a^{x}}{b^{x}}=0$ when $1 < a < b$ | | super-exponential | $x^{x}$ | beats every $a^{x}$ | Reading the table: in a sum, only the fastest-growing term matters; in a quotient, divide by the fastest-growing term present. **Squeeze Theorem (statement with hypotheses).** Suppose $g(x)\le f(x)\le h(x)$ for every $x$ in some interval $(M,\infty)$ (or, for a finite point $a$, for every $x\ne a$ in some interval around $a$), and suppose $\lim g(x)=\lim h(x)=L$ with the **same finite** value $L$. Then $\lim f(x)$ exists and equals $L$. Two hypotheses do the work: the inequality need only hold *eventually*, but the two outer limits must **agree** — if they differ, the theorem says nothing. This is the standard tool for a bounded oscillating numerator, e.g. $-\dfrac{1}{x}\le\dfrac{\sin x}{x}\le\dfrac{1}{x}$ for $x > 0$ forces $\dfrac{\sin x}{x}\to 0$ as $x\to\infty$; l'Hôpital is useless there because $\dfrac{\cos x}{1}$ has no limit. **l'Hôpital's Rule (statement with hypotheses).** Suppose $f$ and $g$ are differentiable on an open interval containing $a$, except possibly at $a$ (for $a=\pm\infty$, on some interval $(M,\infty)$ or $(-\infty,M)$), with $g'(x)\ne 0$ there. Suppose further that either $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$, or $\lim_{x\to a}|g(x)|=\infty$. If $\lim_{x\to a}\dfrac{f'(x)}{g'(x)}$ exists as a finite number or equals $\pm\infty$, then $\lim_{x\to a}\dfrac{f(x)}{g(x)}$ exists and equals it. If $\dfrac{f'}{g'}$ has no limit (it oscillates, say), the rule gives **no information** and you must argue another way. ### Common mistakes | Wrong | Right | | --- | --- | | $\sqrt{x^{2}}=x$, so $\sqrt{9x^{2}+1}\approx 3x$ as $x\to-\infty$ | $\sqrt{x^{2}}$ equals $-x$ when $x < 0$, so $\sqrt{9x^{2}+1}\approx -3x$, a positive number, as $x\to-\infty$ | | $\infty-\infty=0$, so $\sqrt{x^{2}+5x}-x\to 0$ | $\infty-\infty$ is indeterminate; multiply by the conjugate to get $\dfrac{5}{2}$ | | $\dfrac{x+2}{x^{2}-4}$ has vertical asymptotes at $x=\pm 2$ | Cancel first: $\dfrac{1}{x-2}$ for $x\ne -2$; asymptote only at $x=2$, and a **hole** at $x=-2$ | | $\lim_{x\to 2}\dfrac{1}{x-2}=\infty$ | The two one-sided limits are $-\infty$ and $+\infty$, so the two-sided limit does not exist, even in the extended sense | | l'Hôpital on $\dfrac{2x+\sin x}{x+1}$ gives $\dfrac{2+\cos x}{1}$, which has no limit, so the original has none | The form *is* $\infty/\infty$, but the rule's conclusion needs $\lim f'/g'$ to exist; divide by $x$ and squeeze instead to get $2$ | | l'Hôpital applied to $\dfrac{x+1}{x^{2}+3}$ as $x\to 1$ | Not an indeterminate form: just substitute, getting $\dfrac{1}{2}$ | | The graph can never touch $y=L$ | Crossing a horizontal asymptote is allowed and common; the asymptote only controls the behaviour in the limit | | A degree-3 over degree-1 quotient has a slant asymptote | $n-d=2$: the asymptote is a parabola, not a line; there is no linear asymptote |

Key terms

  • limit at infinity
  • end behaviour
  • horizontal asymptote
  • infinite limit
  • vertical asymptote
  • one-sided limit
  • dominant term
  • indeterminate form
  • conjugate multiplication
  • absolute value identity for the square root
  • slant (oblique) asymptote
  • polynomial long division
  • removable discontinuity (hole)
  • growth-rate hierarchy
  • squeeze theorem
  • l'Hopital's Rule

Practice Problems

Evaluate lim⁡x→∞3x2−5x+12x2+7\displaystyle\lim_{x\to\infty}\frac{3x^{2}-5x+1}{2x^{2}+7} and state the horizontal asymptote it produces.

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Do not compare terms by eye: divide the numerator and the denominator by the highest power of xx appearing in the denominator.

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The form is ∞∞\dfrac{\infty}{\infty}, which is indeterminate, so we must rewrite before taking limits.

The highest power in the denominator is x2x^{2}. For x≠0x\ne 0 we may divide top and bottom by x2x^{2} without changing the value:

\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}}} = \frac{3-\dfrac{5}{x}+\dfrac{1}{x^{2}}}{2+\dfrac{7}{x^{2}}}. \end{aligned}$$ As $x\to\infty$ we have $\dfrac{5}{x}\to 0$, $\dfrac{1}{x^{2}}\to 0$ and $\dfrac{7}{x^{2}}\to 0$. The denominator tends to $2\ne 0$, so the quotient law applies: $$\lim_{x\to\infty}\frac{3x^{2}-5x+1}{2x^{2}+7}=\frac{3-0+0}{2+0}=\frac{3}{2}.$$ **Check.** Nothing in the computation used the sign of $x$, so the same work gives $\lim_{x\to-\infty}f(x)=\dfrac{3}{2}$ as well. Numerically, $f(100)=\dfrac{30000-500+1}{20000+7}\approx 1.4745$, consistent with $1.5$. **Answer:** the limit is $\dfrac{3}{2}$, and $y=\dfrac{3}{2}$ is a horizontal asymptote (in both directions).

Evaluate each limit and say what it means for the graph.

(a) lim⁡x→−∞7x−4x2+3\displaystyle\lim_{x\to-\infty}\frac{7x-4}{x^{2}+3} (b) lim⁡x→∞2x3−x5x2+4\displaystyle\lim_{x\to\infty}\frac{2x^{3}-x}{5x^{2}+4}

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In both parts divide by x2x^{2}, the highest power in the denominator, and then look at what the new numerator does.

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(a) Divide numerator and denominator by x2x^{2}: 7x−4x2+3=7x−4x21+3x2.\frac{7x-4}{x^{2}+3}=\frac{\dfrac{7}{x}-\dfrac{4}{x^{2}}}{1+\dfrac{3}{x^{2}}}. As x→−∞x\to-\infty, 7x→0\dfrac{7}{x}\to 0, 4x2→0\dfrac{4}{x^{2}}\to 0, 3x2→0\dfrac{3}{x^{2}}\to 0, so the limit is 0−01+0=0\dfrac{0-0}{1+0}=0.

This is the general rule "numerator degree 11 less than denominator degree 22 gives 00"; the graph has the horizontal asymptote y=0y=0. (Numerical check: at x=−1000x=-1000 the value is −70041000003≈−0.007\dfrac{-7004}{1000003}\approx -0.007.)

(b) Divide numerator and denominator by x2x^{2}: 2x3−x5x2+4=2x−1x5+4x2.\frac{2x^{3}-x}{5x^{2}+4}=\frac{2x-\dfrac{1}{x}}{5+\dfrac{4}{x^{2}}}. As x→∞x\to\infty the denominator tends to 55 while the numerator 2x−1x→∞2x-\dfrac{1}{x}\to\infty. A quantity tending to +∞+\infty divided by one tending to 5>05 > 0 tends to +∞+\infty, so lim⁡x→∞2x3−x5x2+4=+∞.\lim_{x\to\infty}\frac{2x^{3}-x}{5x^{2}+4}=+\infty. There is no horizontal asymptote in this direction. Since the degrees differ by exactly 11, there is a slant asymptote; long division gives 2x3−x5x2+4=25x−135x5x2+4,\frac{2x^{3}-x}{5x^{2}+4}=\frac{2}{5}x-\frac{\frac{13}{5}x}{5x^{2}+4}, because 25x(5x2+4)=2x3+85x\dfrac{2}{5}x\left(5x^{2}+4\right)=2x^{3}+\dfrac{8}{5}x and 2x3−x−(2x3+85x)=−135x2x^{3}-x-\left(2x^{3}+\dfrac{8}{5}x\right)=-\dfrac{13}{5}x. The remainder term tends to 00, so the slant asymptote is y=25xy=\dfrac{2}{5}x.

Answers: (a) 00, horizontal asymptote y=0y=0; (b) +∞+\infty, no horizontal asymptote but the slant asymptote y=25xy=\dfrac{2}{5}x.

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

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Check whether the numerator also vanishes where the denominator does; then decide the sign of the denominator on each side of that point.

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Vertical asymptote. The denominator vanishes only at x=3x=3. There the numerator is 2(3)+1=7≠02(3)+1=7\ne 0, so no cancellation is possible and x=3x=3 is a genuine vertical asymptote (the form is nonzero0\dfrac{\text{nonzero}}{0}).

Signs. Near x=3x=3 the numerator is close to 7>07 > 0.

  • If x→3+x\to 3^{+} then x−3→0x-3\to 0 through positive values, so 2x+1x−3→+∞\dfrac{2x+1}{x-3}\to +\infty.
  • If x→3−x\to 3^{-} then x−3→0x-3\to 0 through negative values, so 2x+1x−3→−∞\dfrac{2x+1}{x-3}\to -\infty.

So lim⁡x→3+f(x)=+∞\lim_{x\to 3^{+}}f(x)=+\infty and lim⁡x→3−f(x)=−∞\lim_{x\to 3^{-}}f(x)=-\infty; the two-sided limit does not exist.

Horizontal asymptote. Divide by xx: f(x)=2+1x1−3x⟶2+01−0=2as x→±∞.f(x)=\frac{2+\dfrac{1}{x}}{1-\dfrac{3}{x}}\longrightarrow \frac{2+0}{1-0}=2 \qquad \text{as } x\to\pm\infty . So y=2y=2 is a horizontal asymptote in both directions.

Independent check by division. 2x+1=2(x−3)+72x+1=2(x-3)+7, hence f(x)=2+7x−3,f(x)=2+\frac{7}{x-3}, which shows at a glance that f(x)→2f(x)\to 2 as x→±∞x\to\pm\infty, that ff blows up at x=3x=3, and that the sign of the blow-up is the sign of x−3x-3. Both methods agree.

Answer: vertical asymptote x=3x=3 with f→+∞f\to+\infty from the right and f→−∞f\to-\infty from the left; horizontal asymptote y=2y=2.

Evaluate lim⁡x→∞2x+sin⁡xx+1\displaystyle\lim_{x\to\infty}\frac{2x+\sin x}{x+1}. Explain why l'Hôpital's Rule cannot be used to finish this problem.

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Divide by xx and deal with the leftover sin⁡xx\dfrac{\sin x}{x} using a bound on sin⁡x\sin x.

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Step 1: divide by xx. For x>0x > 0, 2x+sin⁡xx+1=2+sin⁡xx1+1x.\frac{2x+\sin x}{x+1}=\frac{2+\dfrac{\sin x}{x}}{1+\dfrac{1}{x}}.

Step 2: squeeze the sine term. For all xx we have −1≤sin⁡x≤1-1\le \sin x\le 1, so for x>0x > 0 −1x≤sin⁡xx≤1x.-\frac{1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}. Both outer bounds tend to 00 as x→∞x\to\infty, so by the Squeeze Theorem sin⁡xx→0\dfrac{\sin x}{x}\to 0.

Step 3: combine. The denominator 1+1x→1≠01+\dfrac{1}{x}\to 1\ne 0, so the quotient law gives lim⁡x→∞2x+sin⁡xx+1=2+01=2.\lim_{x\to\infty}\frac{2x+\sin x}{x+1}=\frac{2+0}{1}=2.

Why l'Hôpital fails here. The form is genuinely ∞∞\dfrac{\infty}{\infty} and both functions are differentiable, so the form hypotheses hold. But the rule also requires that lim⁡f′g′\lim \dfrac{f'}{g'} exist (finite or ±∞\pm\infty), and here f′(x)g′(x)=2+cos⁡x1=2+cos⁡x,\frac{f'(x)}{g'(x)}=\frac{2+\cos x}{1}=2+\cos x, which oscillates forever between 11 and 33 and has no limit. When that final hypothesis fails, l'Hôpital's Rule yields no conclusion at all — it does not say the original limit fails to exist. The squeeze argument shows the limit is 22.

Answer: the limit is 22 (so y=2y=2 is a horizontal asymptote), and l'Hôpital is inapplicable because lim⁡(2+cos⁡x)\lim(2+\cos x) does not exist.

Find all horizontal asymptotes of f(x)=9x2+2x4x+3f(x)=\dfrac{\sqrt{9x^{2}+2x}}{4x+3}, being careful with the direction x→−∞x\to-\infty.

Show hint

Write 9x2+2x=x29+2x\sqrt{9x^{2}+2x}=\sqrt{x^{2}}\sqrt{9+\frac{2}{x}} and remember what x2\sqrt{x^{2}} equals when xx is negative.

Show answer

Domain. We need 9x2+2x=x(9x+2)≥09x^{2}+2x=x(9x+2)\ge 0, i.e. x≤−29x\le -\dfrac{2}{9} or x≥0x\ge 0, and x≠−34x\ne -\dfrac{3}{4}. Both tails are in the domain, so both limits make sense.

Factor x2x^{2} out of the root. For every xx in the domain with x≠0x\ne 0, 9x2+2x=x2(9+2x)=x2 9+2x=∣x∣9+2x.\sqrt{9x^{2}+2x}=\sqrt{x^{2}\left(9+\frac{2}{x}\right)}=\sqrt{x^{2}}\,\sqrt{9+\frac{2}{x}}=|x|\sqrt{9+\frac{2}{x}}.

Case x→+∞x\to+\infty. Here ∣x∣=x|x|=x, so dividing numerator and denominator by x>0x > 0, 9x2+2x4x+3=x9+2xx(4+3x)=9+2x4+3x⟶94=34.\frac{\sqrt{9x^{2}+2x}}{4x+3}=\frac{x\sqrt{9+\dfrac{2}{x}}}{x\left(4+\dfrac{3}{x}\right)}=\frac{\sqrt{9+\dfrac{2}{x}}}{4+\dfrac{3}{x}}\longrightarrow \frac{\sqrt{9}}{4}=\frac{3}{4}.

Case x→−∞x\to-\infty. Here ∣x∣=−x|x|=-x, so 9x2+2x4x+3=−x9+2xx(4+3x)=−9+2x4+3x⟶−34.\frac{\sqrt{9x^{2}+2x}}{4x+3}=\frac{-x\sqrt{9+\dfrac{2}{x}}}{x\left(4+\dfrac{3}{x}\right)}=\frac{-\sqrt{9+\dfrac{2}{x}}}{4+\dfrac{3}{x}}\longrightarrow \frac{-3}{4}.

Numerical check. At x=−1000x=-1000: 9x2+2x=8 998 0009x^{2}+2x=8\,998\,000, 8 998 000≈2999.67\sqrt{8\,998\,000}\approx 2999.67, and 4x+3=−39974x+3=-3997, giving ≈−0.7505\approx -0.7505, close to −0.75-0.75. At x=1000x=1000 the value is ≈0.7495\approx 0.7495.

Answer: two horizontal asymptotes, y=34y=\dfrac{3}{4} (as x→+∞x\to+\infty) and y=−34y=-\dfrac{3}{4} (as x→−∞x\to-\infty).

Evaluate

(a) lim⁡x→∞(x2+5x−x)\displaystyle\lim_{x\to\infty}\left(\sqrt{x^{2}+5x}-x\right) (b) lim⁡x→−∞(x2+5x+x)\displaystyle\lim_{x\to-\infty}\left(\sqrt{x^{2}+5x}+x\right)

Show hint

Each is an ∞−∞\infty-\infty form; multiply and divide by the conjugate expression, then divide by xx remembering its sign.

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(a) The form is ∞−∞\infty-\infty, which is indeterminate. Multiply and divide by the conjugate x2+5x+x\sqrt{x^{2}+5x}+x (positive for large xx):

\sqrt{x^{2}+5x}-x &= \frac{\left(\sqrt{x^{2}+5x}-x\right)\left(\sqrt{x^{2}+5x}+x\right)}{\sqrt{x^{2}+5x}+x} = \frac{\left(x^{2}+5x\right)-x^{2}}{\sqrt{x^{2}+5x}+x} = \frac{5x}{\sqrt{x^{2}+5x}+x}. \end{aligned}$$ Now divide top and bottom by $x > 0$, using $x=\sqrt{x^{2}}$ inside the root: $$\frac{5x}{\sqrt{x^{2}+5x}+x}=\frac{5}{\sqrt{1+\dfrac{5}{x}}+1}\longrightarrow \frac{5}{\sqrt{1}+1}=\frac{5}{2}.$$ **(b)** Now $x\to-\infty$ (note $x^{2}+5x\ge 0$ for $x\le -5$, so the expression is defined). Again the form is indeterminate, of type $\infty+(-\infty)$. Multiply and divide by $\sqrt{x^{2}+5x}-x$, which is positive and nonzero here: $$\sqrt{x^{2}+5x}+x=\frac{\left(x^{2}+5x\right)-x^{2}}{\sqrt{x^{2}+5x}-x}=\frac{5x}{\sqrt{x^{2}+5x}-x}.$$ Divide top and bottom by $x$, and this time $x < 0$, so $\dfrac{\sqrt{x^{2}+5x}}{x}=\dfrac{|x|\sqrt{1+\frac{5}{x}}}{x}=-\sqrt{1+\dfrac{5}{x}}$: $$\frac{5x}{\sqrt{x^{2}+5x}-x}=\frac{5}{-\sqrt{1+\dfrac{5}{x}}-1}\longrightarrow \frac{5}{-1-1}=-\frac{5}{2}.$$ **Numerical check of (b).** At $x=-10^{6}$: $\sqrt{x^{2}+5x}\approx 999\,997.5$ and adding $x=-10^{6}$ gives $\approx -2.5$. ✔ **Answers:** (a) $\dfrac{5}{2}$; (b) $-\dfrac{5}{2}$. (Geometrically: $y=\sqrt{x^{2}+5x}$ has the asymptote $y=x+\frac{5}{2}$ on the right and $y=-x-\frac{5}{2}$ on the left.)

Use the growth-rate hierarchy to evaluate

(a) lim⁡x→∞5x2+ex2ex−x3\displaystyle\lim_{x\to\infty}\frac{5x^{2}+e^{x}}{2e^{x}-x^{3}} (b) lim⁡x→∞ln⁡(x2+1)ln⁡(x3+x)\displaystyle\lim_{x\to\infty}\frac{\ln\left(x^{2}+1\right)}{\ln\left(x^{3}+x\right)}

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In (a) divide through by the fastest-growing function present. In (b) pull the dominant power out of each logarithm before dividing.

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(a) The fastest-growing term in sight is exe^{x}, so divide numerator and denominator by exe^{x}: 5x2+ex2ex−x3=5x2e−x+12−x3e−x.\frac{5x^{2}+e^{x}}{2e^{x}-x^{3}}=\frac{5x^{2}e^{-x}+1}{2-x^{3}e^{-x}}. By the hierarchy, lim⁡x→∞xpex=0\lim_{x\to\infty}\dfrac{x^{p}}{e^{x}}=0 for every p>0p > 0, so 5x2e−x→05x^{2}e^{-x}\to 0 and x3e−x→0x^{3}e^{-x}\to 0. The denominator tends to 2≠02\ne 0, so lim⁡x→∞5x2+ex2ex−x3=0+12−0=12.\lim_{x\to\infty}\frac{5x^{2}+e^{x}}{2e^{x}-x^{3}}=\frac{0+1}{2-0}=\frac{1}{2}.

(b) Extract the dominant power inside each logarithm and use ln⁡(AB)=ln⁡A+ln⁡B\ln(AB)=\ln A+\ln B (valid since both factors are positive for x>1x > 1):

\ln\left(x^{2}+1\right) &= \ln\left(x^{2}\left(1+\tfrac{1}{x^{2}}\right)\right)=2\ln x+\ln\left(1+\tfrac{1}{x^{2}}\right),\\ \ln\left(x^{3}+x\right) &= \ln\left(x^{3}\left(1+\tfrac{1}{x^{2}}\right)\right)=3\ln x+\ln\left(1+\tfrac{1}{x^{2}}\right). \end{aligned}$$ Divide top and bottom by $\ln x$ (nonzero for $x > 1$), and note $\ln\left(1+\tfrac{1}{x^{2}}\right)\to \ln 1=0$ while $\ln x\to\infty$, so that ratio tends to $0$: $$\frac{2\ln x+\ln\left(1+\frac{1}{x^{2}}\right)}{3\ln x+\ln\left(1+\frac{1}{x^{2}}\right)}=\frac{2+\dfrac{\ln\left(1+\frac{1}{x^{2}}\right)}{\ln x}}{3+\dfrac{\ln\left(1+\frac{1}{x^{2}}\right)}{\ln x}}\longrightarrow \frac{2}{3}.$$ **Independent check of (b) by l'Hôpital.** The form is $\dfrac{\infty}{\infty}$, both functions are differentiable for $x > 1$, and the denominator's derivative $\dfrac{3x^{2}+1}{x^{3}+x}$ is nonzero there, so the rule may be tried: $$\frac{\dfrac{2x}{x^{2}+1}}{\dfrac{3x^{2}+1}{x^{3}+x}}=\frac{2x\left(x^{3}+x\right)}{\left(x^{2}+1\right)\left(3x^{2}+1\right)}=\frac{2x^{2}\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(3x^{2}+1\right)}=\frac{2x^{2}}{3x^{2}+1}\longrightarrow \frac{2}{3},$$ and since this limit exists, l'Hôpital's conclusion is valid and confirms the answer. **Answers:** (a) $\dfrac{1}{2}$; (b) $\dfrac{2}{3}$.

Let f(x)=x2−4x2−3x+2f(x)=\dfrac{x^{2}-4}{x^{2}-3x+2}. Find all vertical and horizontal asymptotes, identify any removable discontinuity and the value that repairs it, and compute the one-sided limits at each vertical asymptote.

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Factor both quadratics completely before deciding which zeros of the denominator are asymptotes.

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Factor. f(x)=x2−4x2−3x+2=(x−2)(x+2)(x−2)(x−1).f(x)=\frac{x^{2}-4}{x^{2}-3x+2}=\frac{(x-2)(x+2)}{(x-2)(x-1)}. The domain excludes x=2x=2 and x=1x=1.

Cancel. For every x≠2x\ne 2 (and x≠1x\ne 1), f(x)=x+2x−1.f(x)=\frac{x+2}{x-1}.

Removable discontinuity at x=2x=2. The factor (x−2)(x-2) cancelled completely, so x=2x=2 is a hole, not an asymptote: lim⁡x→2f(x)=2+22−1=4.\lim_{x\to 2}f(x)=\frac{2+2}{2-1}=4. Defining f(2)=4f(2)=4 makes ff continuous there.

Vertical asymptote at x=1x=1. After cancellation the denominator still vanishes at x=1x=1 while the numerator equals 1+2=3≠01+2=3\ne 0, so x=1x=1 is a vertical asymptote.

  • As x→1+x\to 1^{+}: numerator →3>0\to 3 > 0, denominator x−1→0+x-1\to 0^{+}, so f(x)→+∞f(x)\to +\infty.
  • As x→1−x\to 1^{-}: numerator →3>0\to 3 > 0, denominator x−1→0−x-1\to 0^{-}, so f(x)→−∞f(x)\to -\infty.

Horizontal asymptote. Using either the original expression or the reduced one, divide by xx: x+2x−1=1+2x1−1x⟶1as x→±∞.\frac{x+2}{x-1}=\frac{1+\dfrac{2}{x}}{1-\dfrac{1}{x}}\longrightarrow 1 \qquad \text{as } x\to\pm\infty . (The degrees of the original numerator and denominator are equal, with leading coefficients 11 and 11, giving the same value.) So y=1y=1 is a horizontal asymptote in both directions.

Answer: hole at (2,4)\left(2,4\right); vertical asymptote x=1x=1 with f→+∞f\to+\infty from the right and f→−∞f\to-\infty from the left; horizontal asymptote y=1y=1.

Find the slant asymptote and the vertical asymptote of f(x)=2x2−3x+1x−2f(x)=\dfrac{2x^{2}-3x+1}{x-2}, and determine on which side of the slant asymptote the curve lies as x→+∞x\to+\infty and as x→−∞x\to-\infty.

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The degree of the numerator exceeds that of the denominator by exactly one: do polynomial long division and look at the remainder term.

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Step 1: long division. Divide 2x2−3x+12x^{2}-3x+1 by x−2x-2:

2x^{2}-3x+1 &= 2x(x-2)+\left(x+1\right)\\ &= 2x(x-2)+1\cdot(x-2)+3\\ &= (2x+1)(x-2)+3 . \end{aligned}$$ Check: $(2x+1)(x-2)+3=2x^{2}-4x+x-2+3=2x^{2}-3x+1$. ✔ Hence $$f(x)=2x+1+\frac{3}{x-2}.$$ **Step 2: the slant asymptote.** Since $\dfrac{3}{x-2}\to 0$ as $x\to\pm\infty$, we have $f(x)-(2x+1)\to 0$, which is exactly the definition of the line $y=2x+1$ being an asymptote. So the slant asymptote is $$y=2x+1 \qquad \text{(in both directions).}$$ There is no horizontal asymptote, since $f(x)\to+\infty$ as $x\to+\infty$ and $f(x)\to-\infty$ as $x\to-\infty$. **Step 3: the vertical asymptote.** The denominator vanishes at $x=2$; the numerator there is $2(4)-6+1=3\ne 0$, so $x=2$ is a vertical asymptote. From $f(x)=2x+1+\dfrac{3}{x-2}$: - $x\to 2^{+}$: $\dfrac{3}{x-2}\to +\infty$, so $f(x)\to +\infty$; - $x\to 2^{-}$: $\dfrac{3}{x-2}\to -\infty$, so $f(x)\to -\infty$. **Step 4: position relative to the line.** The difference is $f(x)-(2x+1)=\dfrac{3}{x-2}$, whose sign is the sign of $x-2$. Therefore the curve lies **above** the line $y=2x+1$ for $x > 2$ (in particular as $x\to+\infty$) and **below** it for $x < 2$ (in particular as $x\to-\infty$). **Answer:** 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 slant asymptote as $x\to+\infty$ and below it as $x\to-\infty$.

Give a complete asymptote analysis of f(x)=x3+2x2−x+1x2−1f(x)=\dfrac{x^{3}+2x^{2}-x+1}{x^{2}-1}: all vertical asymptotes with the four one-sided limits, and the slant asymptote with the side the curve approaches from.

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Divide the cubic by the quadratic first; the quotient and the remainder answer almost every part of the question at once.

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Step 1: long division. Divide x3+2x2−x+1x^{3}+2x^{2}-x+1 by x2−1x^{2}-1:

x^{3}+2x^{2}-x+1-x\left(x^{2}-1\right) &= 2x^{2}+1,\\ 2x^{2}+1-2\left(x^{2}-1\right) &= 3 . \end{aligned}$$ So the quotient is $x+2$ with remainder $3$: $$f(x)=x+2+\frac{3}{x^{2}-1}.$$ Check: $(x+2)\left(x^{2}-1\right)+3=x^{3}-x+2x^{2}-2+3=x^{3}+2x^{2}-x+1$. ✔ **Step 2: vertical asymptotes.** $x^{2}-1=(x-1)(x+1)$ vanishes at $x=\pm 1$. The numerator at $x=1$ is $1+2-1+1=3\ne 0$ and at $x=-1$ is $-1+2+1+1=3\ne 0$, so no cancellation occurs and both $x=1$ and $x=-1$ are vertical asymptotes. Signs come from $\dfrac{3}{(x-1)(x+1)}$ (the polynomial part $x+2$ stays bounded): - $x\to 1^{+}$: $x-1\to 0^{+}$, $x+1\to 2 > 0$, so $x^{2}-1\to 0^{+}$ and $f(x)\to +\infty$. - $x\to 1^{-}$: $x-1\to 0^{-}$, $x+1\to 2 > 0$, so $x^{2}-1\to 0^{-}$ and $f(x)\to -\infty$. - $x\to -1^{-}$: $x-1\to -2 < 0$, $x+1\to 0^{-}$, so $x^{2}-1\to 0^{+}$ and $f(x)\to +\infty$. - $x\to -1^{+}$: $x-1\to -2 < 0$, $x+1\to 0^{+}$, so $x^{2}-1\to 0^{-}$ and $f(x)\to -\infty$. (Sanity check: for $|x| > 1$ we have $x^{2}-1 > 0$ and for $|x| < 1$ we have $x^{2}-1 < 0$, matching the four signs above.) **Step 3: slant asymptote.** Since $\dfrac{3}{x^{2}-1}\to 0$ as $x\to\pm\infty$, $f(x)-(x+2)\to 0$, so $$y=x+2$$ is a slant asymptote in both directions. Because $x^{2}-1 > 0$ once $|x| > 1$, the difference $\dfrac{3}{x^{2}-1}$ is **positive** in both tails: the curve approaches $y=x+2$ **from above** as $x\to+\infty$ and also **from above** as $x\to-\infty$. **Step 4: horizontal asymptotes.** None: $f(x)\to+\infty$ as $x\to+\infty$ and $f(x)\to-\infty$ as $x\to-\infty$ (the degree of the numerator exceeds that of the denominator). **Answer:** vertical asymptotes $x=1$ (with $+\infty$ on the right, $-\infty$ on the left) and $x=-1$ (with $-\infty$ on the right, $+\infty$ on the left); slant asymptote $y=x+2$, approached from above in both directions; no horizontal asymptote.

Let f(x)=x2+4xf(x)=\sqrt{x^{2}+4x}. State the domain, and find the oblique asymptotes as x→+∞x\to+\infty and as x→−∞x\to-\infty using m=lim⁡f(x)xm=\lim \dfrac{f(x)}{x} and b=lim⁡[f(x)−mx]b=\lim\left[f(x)-mx\right].

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Compute the slope first by pulling x2=∣x∣\sqrt{x^{2}}=|x| out of the root, and get the intercept with a conjugate multiplication.

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Domain. x2+4x=x(x+4)≥0x^{2}+4x=x(x+4)\ge 0 exactly when x≤−4x\le -4 or x≥0x\ge 0. Both tails are included, so both asymptotes make sense.

Right-hand asymptote (x→+∞x\to+\infty, so ∣x∣=x|x|=x).

Slope: m=lim⁡x→∞x2+4xx=lim⁡x→∞x1+4xx=lim⁡x→∞1+4x=1.m=\lim_{x\to\infty}\frac{\sqrt{x^{2}+4x}}{x}=\lim_{x\to\infty}\frac{x\sqrt{1+\dfrac{4}{x}}}{x}=\lim_{x\to\infty}\sqrt{1+\frac{4}{x}}=1 . Intercept (an ∞−∞\infty-\infty form, so use the conjugate):

b &= \lim_{x\to\infty}\left(\sqrt{x^{2}+4x}-x\right) = \lim_{x\to\infty}\frac{\left(x^{2}+4x\right)-x^{2}}{\sqrt{x^{2}+4x}+x} = \lim_{x\to\infty}\frac{4x}{x\left(\sqrt{1+\dfrac{4}{x}}+1\right)}\\ &= \lim_{x\to\infty}\frac{4}{\sqrt{1+\dfrac{4}{x}}+1}=\frac{4}{2}=2 . \end{aligned}$$ So the right-hand asymptote is $y=x+2$. **Left-hand asymptote ($x\to-\infty$, so $|x|=-x$).** Slope: $$m=\lim_{x\to-\infty}\frac{\sqrt{x^{2}+4x}}{x}=\lim_{x\to-\infty}\frac{-x\sqrt{1+\dfrac{4}{x}}}{x}=\lim_{x\to-\infty}\left(-\sqrt{1+\frac{4}{x}}\right)=-1 .$$ Intercept: $$\begin{aligned} b &= \lim_{x\to-\infty}\left(\sqrt{x^{2}+4x}+x\right) = \lim_{x\to-\infty}\frac{\left(x^{2}+4x\right)-x^{2}}{\sqrt{x^{2}+4x}-x} = \lim_{x\to-\infty}\frac{4x}{x\left(-\sqrt{1+\dfrac{4}{x}}-1\right)}\\ &= \lim_{x\to-\infty}\frac{4}{-\sqrt{1+\dfrac{4}{x}}-1}=\frac{4}{-2}=-2 . \end{aligned}$$ (In the third equality we divided numerator and denominator by $x < 0$, which turns $\dfrac{\sqrt{x^{2}+4x}}{x}$ into $-\sqrt{1+\frac{4}{x}}$.) So the left-hand asymptote is $y=-x-2$. **Numerical check.** $f(100)=\sqrt{10400}\approx 101.980$ versus $x+2=102$; $f(-100)=\sqrt{9600}\approx 97.980$ versus $-x-2=98$. Both differences are small and shrinking. ✔ **Answer:** domain $x\le -4$ or $x\ge 0$; oblique asymptotes $y=x+2$ as $x\to+\infty$ and $y=-x-2$ as $x\to-\infty$ (the curve lies just below each line). There are no vertical asymptotes: $f$ is continuous on its whole domain.

Find the value of the constant aa for which lim⁡x→∞(x2+ax−x2+1)=3.\lim_{x\to\infty}\left(\sqrt{x^{2}+ax}-\sqrt{x^{2}+1}\right)=3 . Explain why the naive answer "the limit is 00 because both roots behave like xx" is wrong.

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Both terms grow like xx, so this is ∞−∞\infty-\infty: rationalise with the conjugate and keep track of the linear term that survives.

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Step 1: rationalise. For large xx both radicands are positive. Multiply and divide by the conjugate x2+ax+x2+1\sqrt{x^{2}+ax}+\sqrt{x^{2}+1}:

\sqrt{x^{2}+ax}-\sqrt{x^{2}+1} &= \frac{\left(x^{2}+ax\right)-\left(x^{2}+1\right)}{\sqrt{x^{2}+ax}+\sqrt{x^{2}+1}} = \frac{ax-1}{\sqrt{x^{2}+ax}+\sqrt{x^{2}+1}} . \end{aligned}$$ **Step 2: divide by $x$.** For $x > 0$ we have $x=\sqrt{x^{2}}$, so each root divided by $x$ can be written with the $x^{2}$ pulled inside: $$\frac{ax-1}{\sqrt{x^{2}+ax}+\sqrt{x^{2}+1}} =\frac{a-\dfrac{1}{x}}{\sqrt{1+\dfrac{a}{x}}+\sqrt{1+\dfrac{1}{x^{2}}}} .$$ **Step 3: take the limit.** As $x\to\infty$, $\dfrac{1}{x}\to 0$, $\dfrac{a}{x}\to 0$ and $\dfrac{1}{x^{2}}\to 0$, so $$\lim_{x\to\infty}\left(\sqrt{x^{2}+ax}-\sqrt{x^{2}+1}\right)=\frac{a-0}{\sqrt{1}+\sqrt{1}}=\frac{a}{2}.$$ **Step 4: solve.** $\dfrac{a}{2}=3\Rightarrow a=6$. (With $a=6$ the radicand $x^{2}+6x$ is positive for all large $x$, so the expression is defined; the answer is consistent.) **Numerical check.** With $a=6$ and $x=10^{6}$: $\sqrt{10^{12}+6\times 10^{6}}\approx 1\,000\,003.000$ and $\sqrt{10^{12}+1}\approx 1\,000\,000.000$, whose difference is $\approx 3$. ✔ **Why the naive answer fails.** It is true that both roots are asymptotic to $x$, i.e. each equals $x+O(1)$, but the constant-order parts need not cancel: $\sqrt{x^{2}+ax}=x+\dfrac{a}{2}+o(1)$ while $\sqrt{x^{2}+1}=x+o(1)$. Subtracting two quantities that both tend to $\infty$ is an indeterminate form precisely because the difference of the *lower-order* terms survives. Only an algebraic rearrangement (here, the conjugate) makes that surviving term visible. **Answer:** $a=6$.