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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):

3x+22x2+1=3x+2x2x2+1x2=3+2x2+1x2⟶3+02+0=32=322.\begin{aligned} \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→−∞x\to-\infty we have x<0x < 0, so x=−x2x=-\sqrt{x^{2}} and dividing by xx introduces a minus sign inside the root:

3x+22x2+1=3+2x−2+1x2⟶3−2=−322.\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=322y=\dfrac{3\sqrt{2}}{2} and y=−322y=-\dfrac{3\sqrt{2}}{2}.

5. Slant (oblique) asymptotes

Definition. The line y=mx+by=mx+b is an asymptote of ff as x→∞x\to\infty if lim⁡x→∞[f(x)−(mx+b)]=0\lim_{x\to\infty}\left[f(x)-(mx+b)\right]=0. Equivalently, compute m=lim⁡x→∞f(x)xm=\lim_{x\to\infty}\dfrac{f(x)}{x} and then b=lim⁡x→∞[f(x)−mx]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→−∞x\to-\infty, possibly with a different line.

For a rational function with n=d+1n=d+1, polynomial long division gives f(x)=(mx+b)+r(x)q(x)f(x)=(mx+b)+\dfrac{r(x)}{q(x)} with deg r<deg q\text{deg }r < \text{deg }q, so the remainder term tends to 00 and y=mx+by=mx+b is the slant asymptote. The sign of that remainder, r(x)q(x)=f(x)−(mx+b)\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)=x2+1x−1f(x)=\dfrac{x^{2}+1}{x-1}.

Long division: x2+1=(x+1)(x−1)+2x^{2}+1=(x+1)(x-1)+2, so f(x)=x+1+2x−1.f(x)=x+1+\frac{2}{x-1}. Since 2x−1→0\dfrac{2}{x-1}\to 0 as x→±∞x\to\pm\infty, the line y=x+1y=x+1 is a slant asymptote in both directions; the curve lies above it for x>1x > 1 and below it for x<1x < 1. The numerator at x=1x=1 is 2≠02\ne 0, so x=1x=1 is a vertical asymptote, with lim⁡x→1+f(x)=+∞\lim_{x\to 1^{+}}f(x)=+\infty and lim⁡x→1−f(x)=−∞\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 fastestRepresentativeFact (all limits as x→∞x\to\infty)
logarithms(ln⁡x)q(\ln x)^{q}lim⁡(ln⁡x)qxp=0\lim \dfrac{(\ln x)^{q}}{x^{p}}=0 for all p>0p > 0, all qq
powers and rootsxpx^{p}lim⁡xpecx=0\lim \dfrac{x^{p}}{e^{cx}}=0 for all p>0p > 0, c>0c > 0
exponentialsaxa^{x} with a>1a > 1lim⁡axbx=0\lim \dfrac{a^{x}}{b^{x}}=0 when 1<a<b1 < a < b
super-exponentialxxx^{x}beats every axa^{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)≤f(x)≤h(x)g(x)\le f(x)\le h(x) for every xx in some interval (M,∞)(M,\infty) (or, for a finite point aa, for every x≠ax\ne a in some interval around aa), and suppose lim⁡g(x)=lim⁡h(x)=L\lim g(x)=\lim h(x)=L with the same finite value LL. Then lim⁡f(x)\lim f(x) exists and equals LL. 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. −1x≤sin⁡xx≤1x-\dfrac{1}{x}\le\dfrac{\sin x}{x}\le\dfrac{1}{x} for x>0x > 0 forces sin⁡xx→0\dfrac{\sin x}{x}\to 0 as x→∞x\to\infty; l'Hôpital is useless there because cos⁡x1\dfrac{\cos x}{1} has no limit.

l'Hôpital's Rule (statement with hypotheses). Suppose ff and gg are differentiable on an open interval containing aa, except possibly at aa (for a=±∞a=\pm\infty, on some interval (M,∞)(M,\infty) or (−∞,M)(-\infty,M)), with g′(x)≠0g'(x)\ne 0 there. Suppose further that either lim⁡x→af(x)=lim⁡x→ag(x)=0\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0, or lim⁡x→a∣g(x)∣=∞\lim_{x\to a}|g(x)|=\infty. If lim⁡x→af′(x)g′(x)\lim_{x\to a}\dfrac{f'(x)}{g'(x)} exists as a finite number or equals ±∞\pm\infty, then lim⁡x→af(x)g(x)\lim_{x\to a}\dfrac{f(x)}{g(x)} exists and equals it. If f′g′\dfrac{f'}{g'} has no limit (it oscillates, say), the rule gives no information and you must argue another way.

Common mistakes

WrongRight
x2=x\sqrt{x^{2}}=x, so 9x2+1≈3x\sqrt{9x^{2}+1}\approx 3x as x→−∞x\to-\inftyx2\sqrt{x^{2}} equals −x-x when x<0x < 0, so 9x2+1≈−3x\sqrt{9x^{2}+1}\approx -3x, a positive number, as x→−∞x\to-\infty
∞−∞=0\infty-\infty=0, so x2+5x−x→0\sqrt{x^{2}+5x}-x\to 0∞−∞\infty-\infty is indeterminate; multiply by the conjugate to get 52\dfrac{5}{2}
x+2x2−4\dfrac{x+2}{x^{2}-4} has vertical asymptotes at x=±2x=\pm 2Cancel first: 1x−2\dfrac{1}{x-2} for x≠−2x\ne -2; asymptote only at x=2x=2, and a hole at x=−2x=-2
lim⁡x→21x−2=∞\lim_{x\to 2}\dfrac{1}{x-2}=\inftyThe two one-sided limits are −∞-\infty and +∞+\infty, so the two-sided limit does not exist, even in the extended sense
l'Hôpital on 2x+sin⁡xx+1\dfrac{2x+\sin x}{x+1} gives 2+cos⁡x1\dfrac{2+\cos x}{1}, which has no limit, so the original has noneThe form is ∞/∞\infty/\infty, but the rule's conclusion needs lim⁡f′/g′\lim f'/g' to exist; divide by xx and squeeze instead to get 22
l'Hôpital applied to x+1x2+3\dfrac{x+1}{x^{2}+3} as x→1x\to 1Not an indeterminate form: just substitute, getting 12\dfrac{1}{2}
The graph can never touch y=Ly=LCrossing 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 asymptoten−d=2n-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.

Show hint

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

Show answer

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:

3x2−5x+12x2+7=3x2x2−5xx2+1x22x2x2+7x2=3−5x+1x22+7x2.\begin{aligned} \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→∞x\to\infty we have 5x→0\dfrac{5}{x}\to 0, 1x2→0\dfrac{1}{x^{2}}\to 0 and 7x2→0\dfrac{7}{x^{2}}\to 0. The denominator tends to 2≠02\ne 0, so the quotient law applies: lim⁡x→∞3x2−5x+12x2+7=3−0+02+0=32.\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 xx, so the same work gives lim⁡x→−∞f(x)=32\lim_{x\to-\infty}f(x)=\dfrac{3}{2} as well. Numerically, f(100)=30000−500+120000+7≈1.4745f(100)=\dfrac{30000-500+1}{20000+7}\approx 1.4745, consistent with 1.51.5.

Answer: the limit is 32\dfrac{3}{2}, and y=32y=\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}

Show hint

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

Show answer

(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.

Show hint

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

Show answer

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.

Show hint

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

Show answer

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)

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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):

x2+5x−x=(x2+5x−x)(x2+5x+x)x2+5x+x=(x2+5x)−x2x2+5x+x=5xx2+5x+x.\begin{aligned} \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>0x > 0, using x=x2x=\sqrt{x^{2}} inside the root: 5xx2+5x+x=51+5x+1⟶51+1=52.\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→−∞x\to-\infty (note x2+5x≥0x^{2}+5x\ge 0 for x≤−5x\le -5, so the expression is defined). Again the form is indeterminate, of type ∞+(−∞)\infty+(-\infty). Multiply and divide by x2+5x−x\sqrt{x^{2}+5x}-x, which is positive and nonzero here: x2+5x+x=(x2+5x)−x2x2+5x−x=5xx2+5x−x.\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 xx, and this time x<0x < 0, so x2+5xx=∣x∣1+5xx=−1+5x\dfrac{\sqrt{x^{2}+5x}}{x}=\dfrac{|x|\sqrt{1+\frac{5}{x}}}{x}=-\sqrt{1+\dfrac{5}{x}}: 5xx2+5x−x=5−1+5x−1⟶5−1−1=−52.\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=−106x=-10^{6}: x2+5x≈999 997.5\sqrt{x^{2}+5x}\approx 999\,997.5 and adding x=−106x=-10^{6} gives ≈−2.5\approx -2.5. ✔

Answers: (a) 52\dfrac{5}{2}; (b) −52-\dfrac{5}{2}. (Geometrically: y=x2+5xy=\sqrt{x^{2}+5x} has the asymptote y=x+52y=x+\frac{5}{2} on the right and y=−x−52y=-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⁡(x2+1)=ln⁡(x2(1+1x2))=2ln⁡x+ln⁡(1+1x2),ln⁡(x3+x)=ln⁡(x3(1+1x2))=3ln⁡x+ln⁡(1+1x2).\begin{aligned} \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\ln x (nonzero for x>1x > 1), and note ln⁡(1+1x2)→ln⁡1=0\ln\left(1+\tfrac{1}{x^{2}}\right)\to \ln 1=0 while ln⁡x→∞\ln x\to\infty, so that ratio tends to 00: 2ln⁡x+ln⁡(1+1x2)3ln⁡x+ln⁡(1+1x2)=2+ln⁡(1+1x2)ln⁡x3+ln⁡(1+1x2)ln⁡x⟶23.\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>1x > 1, and the denominator's derivative 3x2+1x3+x\dfrac{3x^{2}+1}{x^{3}+x} is nonzero there, so the rule may be tried: 2xx2+13x2+1x3+x=2x(x3+x)(x2+1)(3x2+1)=2x2(x2+1)(x2+1)(3x2+1)=2x23x2+1⟶23,\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) 12\dfrac{1}{2}; (b) 23\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:

2x2−3x+1=2x(x−2)+(x+1)=2x(x−2)+1⋅(x−2)+3=(2x+1)(x−2)+3.\begin{aligned} 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=2x2−4x+x−2+3=2x2−3x+1(2x+1)(x-2)+3=2x^{2}-4x+x-2+3=2x^{2}-3x+1. ✔ Hence f(x)=2x+1+3x−2.f(x)=2x+1+\frac{3}{x-2}.

Step 2: the slant asymptote. Since 3x−2→0\dfrac{3}{x-2}\to 0 as x→±∞x\to\pm\infty, we have f(x)−(2x+1)→0f(x)-(2x+1)\to 0, which is exactly the definition of the line y=2x+1y=2x+1 being an asymptote. So the slant asymptote is y=2x+1(in both directions).y=2x+1 \qquad \text{(in both directions).} There is no horizontal asymptote, since f(x)→+∞f(x)\to+\infty as x→+∞x\to+\infty and f(x)→−∞f(x)\to-\infty as x→−∞x\to-\infty.

Step 3: the vertical asymptote. The denominator vanishes at x=2x=2; the numerator there is 2(4)−6+1=3≠02(4)-6+1=3\ne 0, so x=2x=2 is a vertical asymptote. From f(x)=2x+1+3x−2f(x)=2x+1+\dfrac{3}{x-2}:

  • x→2+x\to 2^{+}: 3x−2→+∞\dfrac{3}{x-2}\to +\infty, so f(x)→+∞f(x)\to +\infty;
  • x→2−x\to 2^{-}: 3x−2→−∞\dfrac{3}{x-2}\to -\infty, so f(x)→−∞f(x)\to -\infty.

Step 4: position relative to the line. The difference is f(x)−(2x+1)=3x−2f(x)-(2x+1)=\dfrac{3}{x-2}, whose sign is the sign of x−2x-2. Therefore the curve lies above the line y=2x+1y=2x+1 for x>2x > 2 (in particular as x→+∞x\to+\infty) and below it for x<2x < 2 (in particular as x→−∞x\to-\infty).

Answer: slant asymptote y=2x+1y=2x+1; vertical asymptote x=2x=2 with f→+∞f\to+\infty from the right and f→−∞f\to-\infty from the left; the curve is above the slant asymptote as x→+∞x\to+\infty and below it as x→−∞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:

x3+2x2−x+1−x(x2−1)=2x2+1,2x2+1−2(x2−1)=3.\begin{aligned} 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+2x+2 with remainder 33: f(x)=x+2+3x2−1.f(x)=x+2+\frac{3}{x^{2}-1}. Check: (x+2)(x2−1)+3=x3−x+2x2−2+3=x3+2x2−x+1(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. x2−1=(x−1)(x+1)x^{2}-1=(x-1)(x+1) vanishes at x=±1x=\pm 1. The numerator at x=1x=1 is 1+2−1+1=3≠01+2-1+1=3\ne 0 and at x=−1x=-1 is −1+2+1+1=3≠0-1+2+1+1=3\ne 0, so no cancellation occurs and both x=1x=1 and x=−1x=-1 are vertical asymptotes.

Signs come from 3(x−1)(x+1)\dfrac{3}{(x-1)(x+1)} (the polynomial part x+2x+2 stays bounded):

  • x→1+x\to 1^{+}: x−1→0+x-1\to 0^{+}, x+1→2>0x+1\to 2 > 0, so x2−1→0+x^{2}-1\to 0^{+} and f(x)→+∞f(x)\to +\infty.
  • x→1−x\to 1^{-}: x−1→0−x-1\to 0^{-}, x+1→2>0x+1\to 2 > 0, so x2−1→0−x^{2}-1\to 0^{-} and f(x)→−∞f(x)\to -\infty.
  • x→−1−x\to -1^{-}: x−1→−2<0x-1\to -2 < 0, x+1→0−x+1\to 0^{-}, so x2−1→0+x^{2}-1\to 0^{+} and f(x)→+∞f(x)\to +\infty.
  • x→−1+x\to -1^{+}: x−1→−2<0x-1\to -2 < 0, x+1→0+x+1\to 0^{+}, so x2−1→0−x^{2}-1\to 0^{-} and f(x)→−∞f(x)\to -\infty.

(Sanity check: for ∣x∣>1|x| > 1 we have x2−1>0x^{2}-1 > 0 and for ∣x∣<1|x| < 1 we have x2−1<0x^{2}-1 < 0, matching the four signs above.)

Step 3: slant asymptote. Since 3x2−1→0\dfrac{3}{x^{2}-1}\to 0 as x→±∞x\to\pm\infty, f(x)−(x+2)→0f(x)-(x+2)\to 0, so y=x+2y=x+2 is a slant asymptote in both directions. Because x2−1>0x^{2}-1 > 0 once ∣x∣>1|x| > 1, the difference 3x2−1\dfrac{3}{x^{2}-1} is positive in both tails: the curve approaches y=x+2y=x+2 from above as x→+∞x\to+\infty and also from above as x→−∞x\to-\infty.

Step 4: horizontal asymptotes. None: f(x)→+∞f(x)\to+\infty as x→+∞x\to+\infty and f(x)→−∞f(x)\to-\infty as x→−∞x\to-\infty (the degree of the numerator exceeds that of the denominator).

Answer: vertical asymptotes x=1x=1 (with +∞+\infty on the right, −∞-\infty on the left) and x=−1x=-1 (with −∞-\infty on the right, +∞+\infty on the left); slant asymptote y=x+2y=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→∞(x2+4x−x)=lim⁡x→∞(x2+4x)−x2x2+4x+x=lim⁡x→∞4xx(1+4x+1)=lim⁡x→∞41+4x+1=42=2.\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}

So the right-hand asymptote is y=x+2y=x+2.

Left-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}\left(-\sqrt{1+\frac{4}{x}}\right)=-1 . Intercept:

b=lim⁡x→−∞(x2+4x+x)=lim⁡x→−∞(x2+4x)−x2x2+4x−x=lim⁡x→−∞4xx(−1+4x−1)=lim⁡x→−∞4−1+4x−1=4−2=−2.\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<0x < 0, which turns x2+4xx\dfrac{\sqrt{x^{2}+4x}}{x} into −1+4x-\sqrt{1+\frac{4}{x}}.) So the left-hand asymptote is y=−x−2y=-x-2.

Numerical check. f(100)=10400≈101.980f(100)=\sqrt{10400}\approx 101.980 versus x+2=102x+2=102; f(−100)=9600≈97.980f(-100)=\sqrt{9600}\approx 97.980 versus −x−2=98-x-2=98. Both differences are small and shrinking. ✔

Answer: domain x≤−4x\le -4 or x≥0x\ge 0; oblique asymptotes y=x+2y=x+2 as x→+∞x\to+\infty and y=−x−2y=-x-2 as x→−∞x\to-\infty (the curve lies just below each line). There are no vertical asymptotes: ff 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}:

x2+ax−x2+1=(x2+ax)−(x2+1)x2+ax+x2+1=ax−1x2+ax+x2+1.\begin{aligned} \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 xx. For x>0x > 0 we have x=x2x=\sqrt{x^{2}}, so each root divided by xx can be written with the x2x^{2} pulled inside:

ax−1x2+ax+x2+1=a−1x1+ax+1+1x2.\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→∞x\to\infty, 1x→0\dfrac{1}{x}\to 0, ax→0\dfrac{a}{x}\to 0 and 1x2→0\dfrac{1}{x^{2}}\to 0, so lim⁡x→∞(x2+ax−x2+1)=a−01+1=a2.\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. a2=3⇒a=6\dfrac{a}{2}=3\Rightarrow a=6. (With a=6a=6 the radicand x2+6xx^{2}+6x is positive for all large xx, so the expression is defined; the answer is consistent.)

Numerical check. With a=6a=6 and x=106x=10^{6}: 1012+6×106≈1 000 003.000\sqrt{10^{12}+6\times 10^{6}}\approx 1\,000\,003.000 and 1012+1≈1 000 000.000\sqrt{10^{12}+1}\approx 1\,000\,000.000, whose difference is ≈3\approx 3. ✔

Why the naive answer fails. It is true that both roots are asymptotic to xx, i.e. each equals x+O(1)x+O(1), but the constant-order parts need not cancel: x2+ax=x+a2+o(1)\sqrt{x^{2}+ax}=x+\dfrac{a}{2}+o(1) while x2+1=x+o(1)\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=6a=6.