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Concept Review

Differential Calculus

L'Hopital's Rule and Indeterminate Forms

Every derivative you have ever computed from the definition was a 00\frac{0}{0} limit, so it is no accident that derivatives can be used to evaluate 00\frac{0}{0} limits. L'Hopital's Rule is that tool: under precise hypotheses it replaces the limit of a quotient by the limit of the quotient of the derivatives. It sits immediately after the differentiation rules and feeds forward into curve sketching and optimisation, where you constantly need the end behaviour of expressions like xln⁡xx\ln x and x2e−xx^{2}e^{-x}. It is also the most over-applied theorem in first-year calculus, so half of this topic is learning when not to reach for it.

What "indeterminate" actually means

A form is indeterminate when the limiting behaviour of the pieces does not by itself determine the limit of the whole. All four limits below are of the form 00\frac{0}{0} as x→0x\to 0, and they do four different things:

lim⁡x→05xx=5,lim⁡x→0x2x=0,lim⁡x→0xx3=+∞,lim⁡x→0xsin⁡1xx does not exist.\lim_{x\to 0}\frac{5x}{x}=5,\qquad \lim_{x\to 0}\frac{x^{2}}{x}=0,\qquad \lim_{x\to 0}\frac{x}{x^{3}}=+\infty,\qquad \lim_{x\to 0}\frac{x\sin\frac{1}{x}}{x}\ \text{does not exist.}

So "00\frac{0}{0}" is not a number and never an answer. It is a report that the limit laws gave no information and that more work is required.

The seven indeterminate forms: 00\dfrac{0}{0}, ∞∞\dfrac{\infty}{\infty}, 0⋅∞0\cdot\infty, ∞−∞\infty-\infty, 1∞1^{\infty}, 000^{0}, ∞0\infty^{0}.

Everything else is determinate: the answer is forced by the pieces, and l'Hopital's Rule must not be used. Here cc denotes a finite nonzero constant.

Form as x→ax\to aForced valueNote
0c\dfrac{0}{c}00just substitute
c0\dfrac{c}{0}infinitethe size blows up; the sign, and whether a two-sided limit exists at all, needs a one-sided look at the denominator — lim⁡x→01x\lim_{x\to 0}\frac{1}{x} is neither +∞+\infty nor −∞-\infty
c∞\dfrac{c}{\infty}00
∞c\dfrac{\infty}{c}infinite
∞+∞\infty+\infty+∞+\inftyonly ∞−∞\infty-\infty is indeterminate
∞⋅∞\infty\cdot\infty+∞+\inftyonly 0⋅∞0\cdot\infty is indeterminate
0+∞0^{+\infty} (base →0+\to 0^{+})00
0−∞0^{-\infty}+∞+\infty
∞+∞\infty^{+\infty}+∞+\infty
∞−∞\infty^{-\infty}00
c∞c^{\infty} with c>1c>1+∞+\inftywith 0<c<10<c<1 the value is 00

Notice how narrow the exponential indeterminates are: only base →1\to 1 with an infinite exponent, base →0+\to 0^{+} with exponent →0\to 0, and base →∞\to\infty with exponent →0\to 0.

L'Hopital's Rule, stated with its hypotheses

Theorem. Let aa be a real number or ±∞\pm\infty, and suppose all four of the following hold.

  1. Differentiability. ff and gg are differentiable on an open interval II containing aa, except possibly at aa itself. (If a=+∞a=+\infty, read this as: differentiable on some interval (N,∞)(N,\infty).) The values f(a)f(a) and g(a)g(a) are irrelevant and need not exist.
  2. Nonvanishing derivative. g′(x)≠0g'(x)\ne 0 for every xx in II with x≠ax\ne a.
  3. The form is indeterminate. Either lim⁡x→af(x)=0\lim_{x\to a}f(x)=0 and lim⁡x→ag(x)=0\lim_{x\to a}g(x)=0, or lim⁡x→af(x)=±∞\lim_{x\to a}f(x)=\pm\infty and lim⁡x→ag(x)=±∞\lim_{x\to a}g(x)=\pm\infty (signs may differ).
  4. The new limit exists. lim⁡x→af′(x)g′(x)=L\lim_{x\to a}\dfrac{f'(x)}{g'(x)}=L, where LL is a real number or ±∞\pm\infty.

Conclusion. lim⁡x→af(x)g(x)=L\lim_{x\to a}\dfrac{f(x)}{g(x)}=L.

The theorem holds verbatim for one-sided limits x→a+x\to a^{+} and x→a−x\to a^{-} (with II a one-sided interval) and for x→−∞x\to-\infty. A refinement worth knowing: in the ∞∞\frac{\infty}{\infty} case it is enough that ∣g(x)∣→∞|g(x)|\to\infty; what ff does is not needed.

Why it is true (the 00\frac{0}{0} case at a finite aa, with ff and gg defined at aa by f(a)=g(a)=0f(a)=g(a)=0, which makes them continuous there). The engine is the Cauchy Mean Value Theorem, and it carries hypotheses of its own.

Cauchy Mean Value Theorem. If ff and gg are continuous on the closed interval [α,β]\left[\alpha,\beta\right], differentiable on the open interval (α,β)\left(\alpha,\beta\right), and g′(t)≠0g'(t)\ne 0 for every tt in (α,β)\left(\alpha,\beta\right), then g(β)≠g(α)g(\beta)\ne g(\alpha) and there is at least one cc in (α,β)\left(\alpha,\beta\right) with

f(β)−f(α)g(β)−g(α)=f′(c)g′(c)\frac{f(\beta)-f(\alpha)}{g(\beta)-g(\alpha)}=\frac{f'(c)}{g'(c)}

Hypotheses 1 and 2 of l'Hopital deliver exactly those conditions on the interval between aa and xx, so for each such xx there is a cc strictly between aa and xx with

f(x)g(x)=f(x)−f(a)g(x)−g(a)=f′(c)g′(c)\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(c)}{g'(c)}

(the middle step is where g(x)≠g(a)=0g(x)\ne g(a)=0 is needed, and hypothesis 2 is what supplies it). As x→ax\to a the point cc is trapped between aa and xx, so c→ac\to a too and the quotient inherits the limit LL.

Every hypothesis earns its place.

  • Drop 3. lim⁡x→0x+2x+1=21=2\lim_{x\to 0}\dfrac{x+2}{x+1}=\dfrac{2}{1}=2 by substitution, but f′g′=11→1\dfrac{f'}{g'}=\dfrac{1}{1}\to 1. Differentiating a quotient that was never indeterminate simply gives the wrong number.
  • Drop 4. lim⁡x→∞x+sin⁡xx=1\lim_{x\to\infty}\dfrac{x+\sin x}{x}=1 (divide by xx), yet f′g′=1+cos⁡x1\dfrac{f'}{g'}=\dfrac{1+\cos x}{1} oscillates and has no limit. When step 4 fails the rule is silent — it does not say the original limit fails to exist.
  • Drop 2. Take f(x)=x+sin⁡xcos⁡xf(x)=x+\sin x\cos x and g(x)=esin⁡xf(x)g(x)=e^{\sin x}f(x) as x→∞x\to\infty. Both tend to +∞+\infty, and since f′(x)=2cos⁡2xf'(x)=2\cos^{2}x and g′(x)=esin⁡xcos⁡x(f(x)+2cos⁡x)g'(x)=e^{\sin x}\cos x\left(f(x)+2\cos x\right), f′(x)g′(x)=2cos⁡xesin⁡x(f(x)+2cos⁡x)→0,\frac{f'(x)}{g'(x)}=\frac{2\cos x}{e^{\sin x}\left(f(x)+2\cos x\right)}\to 0, but f(x)g(x)=e−sin⁡x\dfrac{f(x)}{g(x)}=e^{-\sin x} oscillates between e−1e^{-1} and ee forever. The culprit is that g′(x)=0g'(x)=0 at every x=π2+kπx=\frac{\pi}{2}+k\pi, so no interval (N,∞)(N,\infty) is free of zeros: hypothesis 2 fails, and the conclusion fails with it. (Strictly, f′g′\frac{f'}{g'} is not even defined at those points, so the displayed limit is taken along the set where it does make sense — precisely the loophole hypothesis 2 is written to close.)

Repeated application

If f′g′\dfrac{f'}{g'} is again of type 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}, and f′f' and g′g' themselves satisfy the hypotheses, apply the rule again. Re-verify the form at every single stage. The moment the quotient stops being indeterminate, stop and evaluate.

lim⁡x→0ex−1−xx2 (00) = lim⁡x→0ex−12x (00) = lim⁡x→0ex2=12\lim_{x\to 0}\frac{e^{x}-1-x}{x^{2}}\ \left(\frac{0}{0}\right)\ =\ \lim_{x\to 0}\frac{e^{x}-1}{2x}\ \left(\frac{0}{0}\right)\ =\ \lim_{x\to 0}\frac{e^{x}}{2}=\frac{1}{2}

Growth hierarchy

For every p>0p>0 and every b>1b>1, as x→∞x\to\infty the logarithm grows more slowly than any positive power of xx, which in turn grows more slowly than any exponential:

lim⁡x→∞ln⁡xxp=0andlim⁡x→∞xpbx=0.\lim_{x\to\infty}\frac{\ln x}{x^{p}}=0 \qquad\text{and}\qquad \lim_{x\to\infty}\frac{x^{p}}{b^{x}}=0 .

Both are proved by l'Hopital (the second needs one application for every unit of pp, rounded up). Memorising the hierarchy saves you from grinding out four applications of the rule on x4ex\dfrac{x^{4}}{e^{x}}.

Converting the other five forms into a quotient

FormRecognise it asConvert byYou land on
0⋅∞0\cdot\inftyf⋅gf\cdot g with f→0f\to 0, g→±∞g\to\pm\inftyf1/g\dfrac{f}{1/g} or g1/f\dfrac{g}{1/f}00\frac{0}{0} or ∞∞\frac{\infty}{\infty}
∞−∞\infty-\inftydifference of two blow-upscommon denominator; multiply by the conjugate; factor out the dominant term; or ln⁡A−ln⁡B=ln⁡AB\ln A-\ln B=\ln\frac{A}{B}usually 00\frac{0}{0}
1∞1^{\infty}fgf^{g}, f→1f\to 1, g→±∞g\to\pm\inftytake logarithms0⋅∞0\cdot\infty
000^{0}fgf^{g}, f→0+f\to 0^{+}, g→0g\to 0take logarithms0⋅∞0\cdot\infty
∞0\infty^{0}fgf^{g}, f→∞f\to\infty, g→0g\to 0take logarithms0⋅∞0\cdot\infty

For 0⋅∞0\cdot\infty you have a genuine choice, so pick the arrangement whose derivatives are simpler. Pushing a logarithm into the denominator is almost always the wrong choice: ddx1ln⁡x=−1x(ln⁡x)2\dfrac{d}{dx}\dfrac{1}{\ln x}=\dfrac{-1}{x\left(\ln x\right)^{2}} is worse than what you started with, whereas ddxln⁡x=1x\dfrac{d}{dx}\ln x=\dfrac{1}{x} is a huge simplification. Rule of thumb: differentiate the logarithm, invert the power.

The exponential recipe. To evaluate lim⁡x→af(x)g(x)\lim_{x\to a} f(x)^{g(x)} where f(x)>0f(x)>0 for all xx near aa (this positivity is a genuine requirement — otherwise fgf^{g} need not even be defined):

  1. Set y=f(x)g(x)y=f(x)^{g(x)}, so that ln⁡y=g(x)ln⁡f(x)\ln y=g(x)\ln f(x).
  2. Compute L=lim⁡x→ag(x)ln⁡f(x)L=\lim_{x\to a} g(x)\ln f(x) by turning the 0⋅∞0\cdot\infty form into a quotient.
  3. Since exp⁡\exp is continuous, lim⁡x→ay=eL\lim_{x\to a} y=e^{L}. If L=−∞L=-\infty then the limit is 00; if L=+∞L=+\infty the limit is +∞+\infty.

Step 3 is the step students forget.

When the rule does not apply, or loops

  • The form is not indeterminate. Check first, every time. This is the single most common error.
  • The derivative quotient has no limit. Then the rule tells you nothing; fall back on algebra or the Squeeze Theorem.
  • The rule loops. In lim⁡x→∞x2+1x\lim_{x\to\infty}\dfrac{\sqrt{x^{2}+1}}{x} (form ∞∞\frac{\infty}{\infty}) one application gives x/x2+11=xx2+1\dfrac{x/\sqrt{x^{2}+1}}{1}=\dfrac{x}{\sqrt{x^{2}+1}}; applying the rule to that gives 1x/x2+1=x2+1x\dfrac{1}{x/\sqrt{x^{2}+1}}=\dfrac{\sqrt{x^{2}+1}}{x}, exactly where you started. Divide top and bottom by xx instead: 1+1x2→1\sqrt{1+\frac{1}{x^{2}}}\to 1.
  • The rule makes things worse. Roots, and quotients of polynomials of high degree, usually yield to algebra faster.
  • The rule is circular as a proof. Applying it to lim⁡x→0sin⁡xx\lim_{x\to 0}\frac{\sin x}{x} does return the correct value 11, and once (sin⁡x)′=cos⁡x(\sin x)'=\cos x is on the books that computation is a sound deduction — but you may never cite the rule to establish the limit, because the derivative formula is itself proved from it. The same objection applies to any limit of the form lim⁡h→0f(a+h)−f(a)h\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}: that quotient is the definition of f′(a)f'(a), so differentiating top and bottom assumes what you were asked to find.

Worked example A — an ∞−∞\infty-\infty needing two applications

Evaluate lim⁡x→0+(1x−1sin⁡x)\displaystyle\lim_{x\to 0^{+}}\left(\frac{1}{x}-\frac{1}{\sin x}\right).

As x→0+x\to 0^{+} we get 1x→+∞\frac{1}{x}\to+\infty and 1sin⁡x→+∞\frac{1}{\sin x}\to+\infty: the form is ∞−∞\infty-\infty, which is indeterminate. Combine over a common denominator (legitimate for 0<x<π0<x<\pi, where sin⁡x≠0\sin x\ne 0):

1x−1sin⁡x=sin⁡x−xxsin⁡x\frac{1}{x}-\frac{1}{\sin x}=\frac{\sin x-x}{x\sin x}

Now sin⁡x−x→0\sin x-x\to 0 and xsin⁡x→0x\sin x\to 0, so the form is 00\frac{0}{0}. Both functions are differentiable everywhere, and ddx(xsin⁡x)=sin⁡x+xcos⁡x>0\frac{d}{dx}\left(x\sin x\right)=\sin x+x\cos x>0 for 0<x<π20<x<\frac{\pi}{2}, so hypothesis 2 holds there:

lim⁡x→0+sin⁡x−xxsin⁡x=lim⁡x→0+cos⁡x−1sin⁡x+xcos⁡x\lim_{x\to 0^{+}}\frac{\sin x-x}{x\sin x}=\lim_{x\to 0^{+}}\frac{\cos x-1}{\sin x+x\cos x}

Check the new form before continuing: cos⁡x−1→0\cos x-1\to 0 and sin⁡x+xcos⁡x→0\sin x+x\cos x\to 0, still 00\frac{0}{0}. The next denominator derivative is 2cos⁡x−xsin⁡x2\cos x-x\sin x, which tends to 22 and so is nonzero near 00:

lim⁡x→0+cos⁡x−1sin⁡x+xcos⁡x=lim⁡x→0+−sin⁡x2cos⁡x−xsin⁡x=−02−0=0\lim_{x\to 0^{+}}\frac{\cos x-1}{\sin x+x\cos x}=\lim_{x\to 0^{+}}\frac{-\sin x}{2\cos x-x\sin x}=\frac{-0}{2-0}=0

This last form is 02\frac{0}{2}, not indeterminate, so we stop. Answer: 00. (Sanity check: sin⁡x−x≈−x36\sin x-x\approx-\frac{x^{3}}{6} and xsin⁡x≈x2x\sin x\approx x^{2}, so the quotient behaves like −x6→0-\frac{x}{6}\to 0.)

Worked example B — a 1∞1^{\infty}

Evaluate lim⁡x→0(cos⁡2x)3/x2\displaystyle\lim_{x\to 0}\left(\cos 2x\right)^{3/x^{2}}.

The base cos⁡2x→1\cos 2x\to 1 and the exponent 3x2→+∞\frac{3}{x^{2}}\to+\infty, so the form is 1∞1^{\infty}: indeterminate. For 0<∣x∣<π40<|x|<\frac{\pi}{4} we have cos⁡2x>0\cos 2x>0, so both the power and its logarithm are defined. Put y=(cos⁡2x)3/x2y=\left(\cos 2x\right)^{3/x^{2}}:

ln⁡y=3x2ln⁡(cos⁡2x)=3ln⁡(cos⁡2x)x2\ln y=\frac{3}{x^{2}}\ln\left(\cos 2x\right)=\frac{3\ln\left(\cos 2x\right)}{x^{2}}

The numerator tends to 3ln⁡1=03\ln 1=0 and the denominator to 00: form 00\frac{0}{0}. Differentiate top and bottom separately, using the chain rule on the top and noting g′(x)=2x≠0g'(x)=2x\ne 0 for x≠0x\ne 0:

\lim_{x\to 0}\frac{3\ln\left(\cos 2x\right)}{x^{2}} &= \lim_{x\to 0}\frac{3\cdot\dfrac{-2\sin 2x}{\cos 2x}}{2x}\\ &= \lim_{x\to 0}\frac{-6\tan 2x}{2x}\\ &= \lim_{x\to 0}\left(-3\cdot\frac{\tan 2x}{x}\right)\\ &= -3\cdot 2=-6 \end{aligned}$$ (The last step uses the standard limit $\lim_{x\to 0}\frac{\tan 2x}{x}=2$; one more application of l'Hopital gives the same thing, since $\frac{-12\sec^{2}2x}{2}\to-6$.) So $\ln y\to-6$, and because $\exp$ is continuous, $\lim_{x\to 0} y=e^{-6}$. **Answer: $e^{-6}\approx 0.00248$.** (Sanity check: $\ln\cos 2x\approx-\frac{\left(2x\right)^{2}}{2}=-2x^{2}$, so $\ln y\approx\frac{3\left(-2x^{2}\right)}{x^{2}}=-6$.) ### Common mistakes | Wrong | Right | |---|---| | $\lim_{x\to 0}\frac{\sin x}{x+1}=\lim\frac{\cos x}{1}=1$ | the form is $\frac{0}{1}$, not indeterminate — substitute, and the limit is $0$ | | Differentiating with the Quotient Rule: $\lim\frac{f'g-fg'}{g^{2}}$ | l'Hopital uses $\frac{f'}{g'}$: numerator and denominator are differentiated **separately** | | $\lim_{x\to 0}\frac{e^{x}-1}{x^{2}}=\lim\frac{e^{x}}{2x}=\lim\frac{e^{x}}{2}=\frac{1}{2}$ | after one application the form is $\frac{1}{0}$, which is *not* indeterminate, so you must stop: $\lim_{x\to 0^{+}}=+\infty$, $\lim_{x\to 0^{-}}=-\infty$, and the two-sided limit does not exist | | "$\frac{f'}{g'}$ has no limit, therefore $\frac{f}{g}$ has no limit" | the rule is silent; try algebra instead. $\frac{x+\sin x}{x}\to 1$ even though $\frac{1+\cos x}{1}$ has no limit | | Answering $-6$ for $\lim_{x\to 0}\left(\cos 2x\right)^{3/x^{2}}$ | $-6$ is $\lim\ln y$; the answer is $e^{-6}$ | | Citing the rule as a **proof** that $\lim_{x\to 0}\frac{\sin x}{x}=1$ | circular: $(\sin x)'=\cos x$ is itself proved from this limit, so the rule cannot establish it — the value $1$ it returns is nevertheless correct | | Treating $\frac{1}{0}$, $0\cdot 0$ or $0^{\infty}$ as indeterminate | all determinate: infinite, $0$, and $0$ respectively | | $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=1^{\infty}=1$ | $1^{\infty}$ is indeterminate; the true value is $e$ | | Handling $f^{g}$ with $\left(f^{g}\right)'=gf^{g-1}$ or $f^{g}\ln f$ | neither is valid when both are functions of $x$: take logarithms, or write $f^{g}=e^{g\ln f}$ and use the chain rule | | Applying the rule directly to $\lim_{x\to 0^{+}}x^{x}$ | the theorem is about **quotients**; convert with logarithms first (the answer is $1$) | | Writing $\ln f$ without checking $f>0$ | $\left(\sin x\right)^{x}$ only makes sense where $\sin x>0$, so only $x\to 0^{+}$ is meaningful | | Chasing $\lim_{x\to\infty}\frac{\sqrt{x^{2}+1}}{x}$ with the rule | it loops back to itself; divide by $x$ to get $\sqrt{1+\frac{1}{x^{2}}}\to 1$ | | Quoting the two-sided rule when $f$ or $g$ exists only on one side of $a$ | $\frac{\ln\left(1+x\right)}{\sqrt{x}}$ is defined only for $x>0$, so only $\lim_{x\to 0^{+}}$ can be asked (its value is $0$); use the one-sided version of the theorem |

Key terms

  • Indeterminate form
  • 0/0 form
  • Infinity over infinity form
  • Determinate form
  • L'Hopital's Rule
  • Hypotheses of l'Hopital's Rule
  • Cauchy Mean Value Theorem
  • Repeated application
  • 0 times infinity
  • Infinity minus infinity
  • 1 to the infinity form
  • 0 to the 0 form
  • Infinity to the 0 form
  • Logarithmic conversion
  • Continuity of the exponential function
  • Looping application
  • Growth hierarchy
  • Circular reasoning

Practice Problems

Evaluate lim⁡x→0e2x−1sin⁡3x\displaystyle\lim_{x\to 0}\frac{e^{2x}-1}{\sin 3x}. State the form of the limit and check the hypotheses of l'Hopital's Rule before you use it.

Show hint

Substitute x=0x=0 into the numerator and the denominator separately first — you are not allowed to differentiate until you know what form you are looking at.

Show answer

Step 1 — identify the form. As x→0x\to 0,

e2x−1→e0−1=0,sin⁡3x→sin⁡0=0e^{2x}-1\to e^{0}-1=0,\qquad \sin 3x\to\sin 0=0

so the form is 00\frac{0}{0}: indeterminate, and l'Hopital's Rule is a candidate.

Step 2 — check the remaining hypotheses. Write f(x)=e2x−1f(x)=e^{2x}-1 and g(x)=sin⁡3xg(x)=\sin 3x. Both are differentiable at every real number, so hypothesis 1 holds on any interval around 00. Also g′(x)=3cos⁡3xg'(x)=3\cos 3x, and cos⁡3x≠0\cos 3x\ne 0 whenever ∣x∣<π6|x|<\frac{\pi}{6}, so hypothesis 2 holds on the interval I=(−π6,π6)I=\left(-\frac{\pi}{6},\frac{\pi}{6}\right).

Step 3 — differentiate top and bottom separately.

f′(x)=2e2x,g′(x)=3cos⁡3xf'(x)=2e^{2x},\qquad g'(x)=3\cos 3x

Step 4 — evaluate the new limit. The quotient 2e2x3cos⁡3x\frac{2e^{2x}}{3\cos 3x} is continuous at 00 and its denominator there is 3≠03\ne 0, so direct substitution applies:

lim⁡x→02e2x3cos⁡3x=2e03cos⁡0=23\lim_{x\to 0}\frac{2e^{2x}}{3\cos 3x}=\frac{2e^{0}}{3\cos 0}=\frac{2}{3}

This limit exists, so hypothesis 4 holds and the rule may be applied:

lim⁡x→0e2x−1sin⁡3x=23\lim_{x\to 0}\frac{e^{2x}-1}{\sin 3x}=\frac{2}{3}

Answer: 23\dfrac{2}{3}.

Independent check without l'Hopital. Using the standard limits lim⁡u→0eu−1u=1\lim_{u\to 0}\frac{e^{u}-1}{u}=1 and lim⁡u→0sin⁡uu=1\lim_{u\to 0}\frac{\sin u}{u}=1,

e2x−1sin⁡3x=2x3x⋅e2x−12xsin⁡3x3x→23⋅11=23\frac{e^{2x}-1}{\sin 3x}=\frac{2x}{3x}\cdot\frac{\dfrac{e^{2x}-1}{2x}}{\dfrac{\sin 3x}{3x}}\to\frac{2}{3}\cdot\frac{1}{1}=\frac{2}{3}

Numerically at x=0.001x=0.001 the quotient is 0.002002000.00299999≈0.6673\frac{0.00200200}{0.00299999}\approx 0.6673, consistent with 23\frac{2}{3}.

For each limit below, say whether the form is indeterminate, name the form, and evaluate the limit. Use l'Hopital's Rule only where it is legitimate.

(a) lim⁡x→0sin⁡xx+1\displaystyle\lim_{x\to 0}\frac{\sin x}{x+1}

(b) lim⁡x→∞ln⁡xx\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}

(c) lim⁡x→0+cos⁡xx\displaystyle\lim_{x\to 0^{+}}\frac{\cos x}{x}

(d) lim⁡x→0+x1/x\displaystyle\lim_{x\to 0^{+}}x^{1/x}

(e) lim⁡x→∞(1+3x)x\displaystyle\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{x}

Show hint

For each one, work out separately what the numerator/base does and what the denominator/exponent does, then compare with the list of seven indeterminate forms.

Show answer

(a) Numerator sin⁡x→0\sin x\to 0; denominator x+1→1x+1\to 1. The form is 01\frac{0}{1}, which is not indeterminate. The quotient is continuous at 00 (denominator nonzero), so substitute:

lim⁡x→0sin⁡xx+1=01=0\lim_{x\to 0}\frac{\sin x}{x+1}=\frac{0}{1}=0

Applying l'Hopital here would give cos⁡x1→1\frac{\cos x}{1}\to 1, which is simply wrong.

(b) Both ln⁡x→∞\ln x\to\infty and x→∞x\to\infty: form ∞∞\frac{\infty}{\infty}, indeterminate. On (0,∞)(0,\infty) both functions are differentiable and g′(x)=1≠0g'(x)=1\ne 0, so the rule applies:

lim⁡x→∞ln⁡xx=lim⁡x→∞1/x1=lim⁡x→∞1x=0\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{1/x}{1}=\lim_{x\to\infty}\frac{1}{x}=0

(c) Numerator cos⁡x→1\cos x\to 1; denominator x→0+x\to 0^{+}. The form is 10+\frac{1}{0^{+}}, which is not indeterminate — it is determinate and infinite:

lim⁡x→0+cos⁡xx=+∞\lim_{x\to 0^{+}}\frac{\cos x}{x}=+\infty

(At x=0.001x=0.001 the quotient is about 10001000.) L'Hopital would give −sin⁡x1→0\frac{-\sin x}{1}\to 0: badly wrong.

(d) Base x→0+x\to 0^{+}, exponent 1x→+∞\frac{1}{x}\to+\infty: the form is 0+∞0^{+\infty}, which is not indeterminate. To see the value, note that for x>0x>0

ln⁡(x1/x)=ln⁡xx=(ln⁡x)⋅1x→(−∞)⋅(+∞)=−∞\ln\left(x^{1/x}\right)=\frac{\ln x}{x}=\left(\ln x\right)\cdot\frac{1}{x}\to\left(-\infty\right)\cdot\left(+\infty\right)=-\infty

which is a determinate product. Since exp⁡\exp is continuous, lim⁡x→0+x1/x=e−∞=0\lim_{x\to 0^{+}}x^{1/x}=e^{-\infty}=0.

(e) Base 1+3x→11+\frac{3}{x}\to 1, exponent x→∞x\to\infty: form 1∞1^{\infty}, indeterminate. For x>0x>0 the base is positive, so put y=(1+3x)xy=\left(1+\frac{3}{x}\right)^{x} and take logarithms:

ln⁡y=xln⁡(1+3x)\ln y=x\ln\left(1+\frac{3}{x}\right)

This is ∞⋅0\infty\cdot 0. Substitute t=1xt=\frac{1}{x}, so t→0+t\to 0^{+}:

ln⁡y=ln⁡(1+3t)t\ln y=\frac{\ln\left(1+3t\right)}{t}

Numerator →ln⁡1=0\to\ln 1=0 and denominator →0\to 0: form 00\frac{0}{0}. Both functions are differentiable near 00 and the denominator's derivative is 1≠01\ne 0, so

lim⁡t→0+ln⁡(1+3t)t=lim⁡t→0+31+3t1=3\lim_{t\to 0^{+}}\frac{\ln\left(1+3t\right)}{t}=\lim_{t\to 0^{+}}\frac{\dfrac{3}{1+3t}}{1}=3

Hence ln⁡y→3\ln y\to 3 and lim⁡x→∞(1+3x)x=e3\lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{x}=e^{3}.

Summary: (a) not indeterminate, 00; (b) ∞∞\frac{\infty}{\infty}, 00; (c) not indeterminate, +∞+\infty; (d) not indeterminate (0∞0^{\infty}), 00; (e) 1∞1^{\infty}, e3e^{3}.

Evaluate lim⁡x→∞3x2−5x2x2+7\displaystyle\lim_{x\to\infty}\frac{3x^{2}-5x}{2x^{2}+7} using l'Hopital's Rule, checking the form at every stage. Then confirm your answer by an algebraic method that does not use the rule.

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One application will not finish the job — but before you apply it a second time, look again at what the new numerator and new denominator are doing.

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Stage 0 — the form. As x→∞x\to\infty, 3x2−5x→+∞3x^{2}-5x\to+\infty and 2x2+7→+∞2x^{2}+7\to+\infty: form ∞∞\frac{\infty}{\infty}, indeterminate. Both functions are polynomials, hence differentiable everywhere.

Stage 1. With f(x)=3x2−5xf(x)=3x^{2}-5x and g(x)=2x2+7g(x)=2x^{2}+7 we get f′(x)=6x−5f'(x)=6x-5 and g′(x)=4xg'(x)=4x, and g′(x)≠0g'(x)\ne 0 for all x>0x>0, so hypothesis 2 holds on (0,∞)(0,\infty):

lim⁡x→∞3x2−5x2x2+7=lim⁡x→∞6x−54xprovided the right-hand limit exists.\lim_{x\to\infty}\frac{3x^{2}-5x}{2x^{2}+7}=\lim_{x\to\infty}\frac{6x-5}{4x}\quad\text{provided the right-hand limit exists.}

Stage 2 — re-check the form. Now 6x−5→+∞6x-5\to+\infty and 4x→+∞4x\to+\infty: still ∞∞\frac{\infty}{\infty}, so we may go again. The new derivatives are 66 and 44, and 4≠04\ne 0:

lim⁡x→∞6x−54x=lim⁡x→∞64=32\lim_{x\to\infty}\frac{6x-5}{4x}=\lim_{x\to\infty}\frac{6}{4}=\frac{3}{2}

This limit exists, which validates stage 2, which in turn validates stage 1. Therefore

lim⁡x→∞3x2−5x2x2+7=32\lim_{x\to\infty}\frac{3x^{2}-5x}{2x^{2}+7}=\frac{3}{2}

Answer: 32\dfrac{3}{2}.

Independent check by algebra. Divide numerator and denominator by x2x^{2} (legitimate for x≠0x\ne 0):

3x2−5x2x2+7=3−5x2+7x2→3−02+0=32\frac{3x^{2}-5x}{2x^{2}+7}=\frac{3-\dfrac{5}{x}}{2+\dfrac{7}{x^{2}}}\to\frac{3-0}{2+0}=\frac{3}{2}

Numerically at x=1000x=1000: 29950002000007≈1.4975\frac{2995000}{2000007}\approx 1.4975, consistent with 32\frac{3}{2}. Note the shortcut this illustrates: for a rational function, the limit at infinity is the ratio of the leading coefficients when the degrees match — much faster than the rule.

Evaluate lim⁡x→0+x2ln⁡x\displaystyle\lim_{x\to 0^{+}}x^{2}\ln x. There are two ways to write this 0⋅∞0\cdot\infty product as a quotient; explain which one you choose and why the other is a bad idea.

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A product fgfg can be written as f1/g\frac{f}{1/g} or as g1/f\frac{g}{1/f}. Ask yourself which of ln⁡x\ln x and x2x^{2} you would rather differentiate, and which you would rather push into a denominator.

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Step 1 — identify the form. As x→0+x\to 0^{+}: x2→0x^{2}\to 0 and ln⁡x→−∞\ln x\to-\infty. The form is 0⋅(−∞)0\cdot\left(-\infty\right), which is indeterminate. The rule applies only to quotients, so the product must be rewritten.

Step 2 — choose the rearrangement. The two options are

ln⁡xx−2andx21ln⁡x\frac{\ln x}{x^{-2}}\qquad\text{and}\qquad \frac{x^{2}}{\dfrac{1}{\ln x}}

Choose the first. Differentiating ln⁡x\ln x gives the very simple 1x\frac{1}{x}, and differentiating the power x−2x^{-2} keeps a power. The second option is a bad idea because

ddx(1ln⁡x)=−1x(ln⁡x)2\frac{d}{dx}\left(\frac{1}{\ln x}\right)=\frac{-1}{x\left(\ln x\right)^{2}}

so the derivative quotient becomes 2x−1x(ln⁡x)2=−2x2(ln⁡x)2\dfrac{2x}{-\dfrac{1}{x\left(\ln x\right)^{2}}}=-2x^{2}\left(\ln x\right)^{2}, which is more complicated than the original — the same indeterminate form with the logarithm now squared.

Step 3 — check the hypotheses. On (0,1)\left(0,1\right) both f(x)=ln⁡xf(x)=\ln x and g(x)=x−2g(x)=x^{-2} are differentiable, and g′(x)=−2x−3≠0g'(x)=-2x^{-3}\ne 0 there. The form of ln⁡xx−2\frac{\ln x}{x^{-2}} is −∞+∞\frac{-\infty}{+\infty}, which is the ∞∞\frac{\infty}{\infty} case.

Step 4 — apply the rule and simplify.

\lim_{x\to 0^{+}}\frac{\ln x}{x^{-2}} &= \lim_{x\to 0^{+}}\frac{\dfrac{1}{x}}{-2x^{-3}}\\ &= \lim_{x\to 0^{+}}\frac{1}{x}\cdot\frac{x^{3}}{-2}\\ &= \lim_{x\to 0^{+}}\left(-\frac{x^{2}}{2}\right)\\ &= 0 \end{aligned}$$ The final expression is a polynomial, so no indeterminacy remains and we stop. **Answer: $\displaystyle\lim_{x\to 0^{+}}x^{2}\ln x=0$.** *Independent check.* At $x=10^{-3}$: $x^{2}\ln x=10^{-6}\times\left(-6.9078\right)\approx-6.9\times 10^{-6}$, and at $x=10^{-6}$ it is about $-1.4\times 10^{-11}$ — collapsing to $0$. This is the general fact that a positive power of $x$ beats $\ln x$ as $x\to 0^{+}$, the mirror image of the growth hierarchy at infinity, where $x^{p}$ outgrows $\ln x$.

Evaluate lim⁡x→0x−sin⁡xx3\displaystyle\lim_{x\to 0}\frac{x-\sin x}{x^{3}}, justifying the form at every stage and stating where you could stop early.

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Each application removes only one power of xx from the denominator, so expect to go round more than once — and check the form each time before you go again.

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Stage 0 — the form. x−sin⁡x→0−0=0x-\sin x\to 0-0=0 and x3→0x^{3}\to 0: form 00\frac{0}{0}. Both functions are differentiable everywhere, and g′(x)=3x2≠0g'(x)=3x^{2}\ne 0 for every x≠0x\ne 0, so hypotheses 1 and 2 hold on any interval around 00.

Stage 1.

lim⁡x→0x−sin⁡xx3=lim⁡x→01−cos⁡x3x2\lim_{x\to 0}\frac{x-\sin x}{x^{3}}=\lim_{x\to 0}\frac{1-\cos x}{3x^{2}}

Re-check: 1−cos⁡x→01-\cos x\to 0 and 3x2→03x^{2}\to 0, still 00\frac{0}{0}; the new denominator derivative 6x6x is nonzero for x≠0x\ne 0.

Stage 2.

lim⁡x→01−cos⁡x3x2=lim⁡x→0sin⁡x6x\lim_{x\to 0}\frac{1-\cos x}{3x^{2}}=\lim_{x\to 0}\frac{\sin x}{6x}

You may stop here, because lim⁡x→0sin⁡xx=1\lim_{x\to 0}\frac{\sin x}{x}=1 is a standard limit:

lim⁡x→0sin⁡x6x=16lim⁡x→0sin⁡xx=16\lim_{x\to 0}\frac{\sin x}{6x}=\frac{1}{6}\lim_{x\to 0}\frac{\sin x}{x}=\frac{1}{6}

Stage 3 (if you prefer to continue). The form of sin⁡x6x\frac{\sin x}{6x} is again 00\frac{0}{0}, and the denominator derivative is 6≠06\ne 0, so a third application is permitted:

lim⁡x→0sin⁡x6x=lim⁡x→0cos⁡x6=16\lim_{x\to 0}\frac{\sin x}{6x}=\lim_{x\to 0}\frac{\cos x}{6}=\frac{1}{6}

Read this stage as bookkeeping, not as a proof: it differentiates sin⁡x\sin x, and (sin⁡x)′=cos⁡x(\sin x)'=\cos x is itself derived from lim⁡x→0sin⁡xx=1\lim_{x\to 0}\frac{\sin x}{x}=1, so stage 3 may never be used to establish that standard limit. Stage 2 is the honest place to stop. Either way the limit exists, and working backwards validates stage 2, then stage 1. Therefore

lim⁡x→0x−sin⁡xx3=16\lim_{x\to 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6}

Answer: 16\dfrac{1}{6}.

Independent check. The Maclaurin series gives sin⁡x=x−x36+x5120−⋯\sin x=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\cdots, so

x−sin⁡xx3=x36−x5120+⋯x3=16−x2120+⋯→16\frac{x-\sin x}{x^{3}}=\frac{\dfrac{x^{3}}{6}-\dfrac{x^{5}}{120}+\cdots}{x^{3}}=\frac{1}{6}-\frac{x^{2}}{120}+\cdots\to\frac{1}{6}

Numerically at x=0.1x=0.1: 0.1−0.09983340.001≈0.16658\frac{0.1-0.0998334}{0.001}\approx 0.16658, and 16=0.16667\frac{1}{6}=0.16667.

Evaluate lim⁡x→1(xx−1−1ln⁡x)\displaystyle\lim_{x\to 1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right).

Show hint

Each piece blows up, so the limit laws cannot be used termwise. Put the two fractions over a single denominator first and see what form appears.

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Step 1 — identify the form. As x→1+x\to 1^{+}: x−1→0+x-1\to 0^{+} so xx−1→+∞\frac{x}{x-1}\to+\infty, and ln⁡x→0+\ln x\to 0^{+} so 1ln⁡x→+∞\frac{1}{\ln x}\to+\infty. The form is ∞−∞\infty-\infty, indeterminate (from the left both pieces tend to −∞-\infty, giving −∞+∞-\infty+\infty, equally indeterminate). You may not split the limit across the subtraction, because neither piece has a limit.

Step 2 — make it a single quotient. For x>0x>0 with x≠1x\ne 1 both denominators are nonzero, so

xx−1−1ln⁡x=xln⁡x−(x−1)(x−1)ln⁡x=xln⁡x−x+1(x−1)ln⁡x\frac{x}{x-1}-\frac{1}{\ln x}=\frac{x\ln x-\left(x-1\right)}{\left(x-1\right)\ln x}=\frac{x\ln x-x+1}{\left(x-1\right)\ln x}

Set N(x)=xln⁡x−x+1N(x)=x\ln x-x+1 and D(x)=(x−1)ln⁡xD(x)=\left(x-1\right)\ln x. Then N(1)=0−1+1=0N(1)=0-1+1=0 and D(1)=0D(1)=0, and both are continuous at 11, so the form is 00\frac{0}{0}.

Step 3 — first application. Both NN and DD are differentiable on (0,∞)(0,\infty):

N′(x)=ln⁡x+x⋅1x−1=ln⁡x,D′(x)=ln⁡x+x−1x=ln⁡x+1−1xN'(x)=\ln x+x\cdot\frac{1}{x}-1=\ln x,\qquad D'(x)=\ln x+\frac{x-1}{x}=\ln x+1-\frac{1}{x}

Is D′(x)≠0D'(x)\ne 0 near 11? We have D′′(x)=1x+1x2>0D''(x)=\frac{1}{x}+\frac{1}{x^{2}}>0 for x>0x>0, so D′D' is strictly increasing; since D′(1)=0+1−1=0D'(1)=0+1-1=0, it follows that D′(x)≠0D'(x)\ne 0 for every x≠1x\ne 1. Hypothesis 2 holds. Hence

lim⁡x→1N(x)D(x)=lim⁡x→1ln⁡xln⁡x+1−1x\lim_{x\to 1}\frac{N(x)}{D(x)}=\lim_{x\to 1}\frac{\ln x}{\ln x+1-\dfrac{1}{x}}

Step 4 — re-check the form. Numerator →ln⁡1=0\to\ln 1=0; denominator →0+1−1=0\to 0+1-1=0. Still 00\frac{0}{0}, so apply again. The new derivatives are

ddxln⁡x=1x,ddx(ln⁡x+1−1x)=1x+1x2\frac{d}{dx}\ln x=\frac{1}{x},\qquad \frac{d}{dx}\left(\ln x+1-\frac{1}{x}\right)=\frac{1}{x}+\frac{1}{x^{2}}

and the second is nonzero for all x>0x>0.

Step 5 — finish.

lim⁡x→11x1x+1x2=11+1=12\lim_{x\to 1}\frac{\dfrac{1}{x}}{\dfrac{1}{x}+\dfrac{1}{x^{2}}}=\frac{1}{1+1}=\frac{1}{2}

The last expression is continuous at x=1x=1 with nonzero denominator, so no indeterminacy remains and we stop. Working backwards, both applications were valid, so

lim⁡x→1(xx−1−1ln⁡x)=12\lim_{x\to 1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right)=\frac{1}{2}

Answer: 12\dfrac{1}{2}.

Independent numerical check. At x=1.01x=1.01: 1.010.01=101\frac{1.01}{0.01}=101 and 1ln⁡1.01=10.0099503=100.4992\frac{1}{\ln 1.01}=\frac{1}{0.0099503}=100.4992, difference ≈0.5008\approx 0.5008. At x=0.99x=0.99: 0.99−0.01=−99\frac{0.99}{-0.01}=-99 and 1ln⁡0.99=1−0.0100503=−99.4992\frac{1}{\ln 0.99}=\frac{1}{-0.0100503}=-99.4992, difference ≈0.4992\approx 0.4992. Both sides approach 12\frac{1}{2}.

Evaluate lim⁡x→∞(x+2x−1)3x\displaystyle\lim_{x\to\infty}\left(\frac{x+2}{x-1}\right)^{3x}.

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Identify what the base and the exponent each tend to, then take the natural logarithm of the expression before you do anything else. Remember to undo the logarithm at the end.

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Step 1 — identify the form. For x>1x>1 the base is positive, and

x+2x−1=1+2x1−1x→1,3x→+∞\frac{x+2}{x-1}=\frac{1+\dfrac{2}{x}}{1-\dfrac{1}{x}}\to 1,\qquad 3x\to+\infty

so the form is 1∞1^{\infty}: indeterminate. (Do not write 1∞=11^{\infty}=1.)

Step 2 — take logarithms. Put y=(x+2x−1)3xy=\left(\dfrac{x+2}{x-1}\right)^{3x}, valid for x>1x>1 where the base is positive. Then

ln⁡y=3xln⁡(x+2x−1)=3x[ln⁡(x+2)−ln⁡(x−1)]\ln y=3x\ln\left(\frac{x+2}{x-1}\right)=3x\left[\ln\left(x+2\right)-\ln\left(x-1\right)\right]

This has form ∞⋅0\infty\cdot 0, still indeterminate, so it must be made into a quotient.

Step 3 — make a quotient. Move the factor 3x3x downstairs:

ln⁡y=ln⁡(x+2)−ln⁡(x−1)13x\ln y=\frac{\ln\left(x+2\right)-\ln\left(x-1\right)}{\dfrac{1}{3x}}

Numerator →ln⁡1=0\to\ln 1=0 and denominator →0\to 0, so this is 00\frac{0}{0}. Both functions are differentiable on (1,∞)(1,\infty), and the denominator's derivative −13x2-\frac{1}{3x^{2}} is nonzero there, so hypotheses 1 and 2 hold.

Step 4 — apply the rule.

\frac{d}{dx}\left[\ln\left(x+2\right)-\ln\left(x-1\right)\right] &= \frac{1}{x+2}-\frac{1}{x-1}\\ &= \frac{\left(x-1\right)-\left(x+2\right)}{\left(x+2\right)\left(x-1\right)}\\ &= \frac{-3}{\left(x+2\right)\left(x-1\right)} \end{aligned}$$ and $\dfrac{d}{dx}\left(\dfrac{1}{3x}\right)=-\dfrac{1}{3x^{2}}$. Therefore $$\begin{aligned} \lim_{x\to\infty}\ln y &= \lim_{x\to\infty}\frac{\dfrac{-3}{\left(x+2\right)\left(x-1\right)}}{-\dfrac{1}{3x^{2}}}\\ &= \lim_{x\to\infty}\frac{-3}{\left(x+2\right)\left(x-1\right)}\cdot\left(-3x^{2}\right)\\ &= \lim_{x\to\infty}\frac{9x^{2}}{\left(x+2\right)\left(x-1\right)}\\ &= \lim_{x\to\infty}\frac{9}{\left(1+\dfrac{2}{x}\right)\left(1-\dfrac{1}{x}\right)}\\ &= \frac{9}{1\cdot 1}=9 \end{aligned}$$ **Step 5 — undo the logarithm.** Since $\exp$ is continuous and $\ln y\to 9$, $$\lim_{x\to\infty}\left(\frac{x+2}{x-1}\right)^{3x}=e^{9}$$ **Answer: $e^{9}\approx 8103.1$.** *Independent check.* Write $\frac{x+2}{x-1}=1+\frac{3}{x-1}$, so the expression is $\left(1+\frac{3}{x-1}\right)^{3x}$. Using $\left(1+\frac{a}{n}\right)^{n}\to e^{a}$, the exponent behaves like $3x\cdot\frac{3}{x-1}\to 9$, giving $e^{9}$. Numerically at $x=1000$: $\ln y=3000\ln\left(\frac{1002}{999}\right)=3000\times 0.0029985=8.9955$, and $e^{8.9955}\approx 8067$, closing in on $e^{9}\approx 8103$.

Evaluate lim⁡x→0+(sin⁡x)x\displaystyle\lim_{x\to 0^{+}}\left(\sin x\right)^{x}. Explain why the limit is only posed as a right-hand limit.

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Name the form first. Then take logarithms and choose which of the two factors to send into the denominator — the one whose reciprocal is easy to differentiate.

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Step 1 — why one-sided. The power aba^{b} with a variable exponent is defined by ab=ebln⁡aa^{b}=e^{b\ln a}, which requires a>0a>0. Since sin⁡x>0\sin x>0 exactly for 0<x<π0<x<\pi (near the origin), the expression (sin⁡x)x\left(\sin x\right)^{x} is undefined for x<0x<0 close to 00. Only the right-hand limit can be asked.

Step 2 — identify the form. As x→0+x\to 0^{+}: the base sin⁡x→0+\sin x\to 0^{+} and the exponent x→0x\to 0. The form is 000^{0}: indeterminate.

Step 3 — take logarithms. Let y=(sin⁡x)xy=\left(\sin x\right)^{x} for 0<x<π0<x<\pi. Then

ln⁡y=xln⁡(sin⁡x)\ln y=x\ln\left(\sin x\right)

which has form 0⋅(−∞)0\cdot\left(-\infty\right): still indeterminate.

Step 4 — make a quotient. Keep the logarithm upstairs (it differentiates well) and send xx downstairs:

ln⁡y=ln⁡(sin⁡x)1x\ln y=\frac{\ln\left(\sin x\right)}{\dfrac{1}{x}}

As x→0+x\to 0^{+} the numerator →−∞\to-\infty and the denominator →+∞\to+\infty, so this is the ∞∞\frac{\infty}{\infty} case. On (0,π2)\left(0,\frac{\pi}{2}\right) both functions are differentiable, and the denominator's derivative −1x2-\frac{1}{x^{2}} is never zero there.

Step 5 — apply the rule.

\lim_{x\to 0^{+}}\frac{\ln\left(\sin x\right)}{\dfrac{1}{x}} &= \lim_{x\to 0^{+}}\frac{\dfrac{\cos x}{\sin x}}{-\dfrac{1}{x^{2}}}\\ &= \lim_{x\to 0^{+}}\left(-\frac{x^{2}\cos x}{\sin x}\right)\\ &= \lim_{x\to 0^{+}}\left(-x\cos x\right)\cdot\frac{x}{\sin x} \end{aligned}$$ Now $-x\cos x\to 0$ and $\frac{x}{\sin x}\to 1$ (the reciprocal of the standard limit, allowed because the limit $1$ is nonzero), so by the Product Law $$\lim_{x\to 0^{+}}\ln y=0\cdot 1=0$$ **Step 6 — undo the logarithm.** By continuity of $\exp$, $$\lim_{x\to 0^{+}}\left(\sin x\right)^{x}=e^{0}=1$$ **Answer: $1$.** *Independent check.* $\ln y=x\ln\left(\sin x\right)=x\ln\left(\dfrac{\sin x}{x}\right)+x\ln x$. The first term tends to $0\cdot\ln 1=0$ and the second is the standard limit $x\ln x\to 0$, so $\ln y\to 0$ and $y\to 1$. Numerically at $x=0.01$: $\ln y=0.01\times\ln\left(0.0099998\right)\approx-0.04605$, so $y\approx 0.955$ — creeping up towards $1$ (the approach is slow because $x\ln x$ decays slowly).

Show that repeated use of l'Hopital's Rule on lim⁡x→∞4x2+13x+5\displaystyle\lim_{x\to\infty}\frac{\sqrt{4x^{2}+1}}{3x+5} produces an endless loop, and then evaluate the limit correctly.

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Apply the rule once, simplify the result into a single quotient, and then apply it to that quotient — compare what you get with where you started.

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Step 1 — the form is genuinely indeterminate. As x→∞x\to\infty, 4x2+1→∞\sqrt{4x^{2}+1}\to\infty and 3x+5→∞3x+5\to\infty: form ∞∞\frac{\infty}{\infty}. So the rule is at least allowed; the trouble is that it is useless here.

Step 2 — first application. With f(x)=(4x2+1)1/2f(x)=\left(4x^{2}+1\right)^{1/2} and g(x)=3x+5g(x)=3x+5,

f′(x)=12(4x2+1)−1/2⋅8x=4x4x2+1,g′(x)=3≠0f'(x)=\frac{1}{2}\left(4x^{2}+1\right)^{-1/2}\cdot 8x=\frac{4x}{\sqrt{4x^{2}+1}},\qquad g'(x)=3\ne 0

so the rule gives

lim⁡x→∞4x2+13x+5=lim⁡x→∞4x34x2+1\lim_{x\to\infty}\frac{\sqrt{4x^{2}+1}}{3x+5}=\lim_{x\to\infty}\frac{4x}{3\sqrt{4x^{2}+1}}

provided the right-hand limit exists — and it does: it equals 23\frac{2}{3}, by the same division-by-xx that Step 4 performs. So this first application is perfectly valid; the trap is what happens if you refuse to evaluate and reach for the rule again just because the new quotient still looks like ∞∞\frac{\infty}{\infty}.

Step 3 — second application. The new quotient is again ∞∞\frac{\infty}{\infty}. Differentiating numerator 4x4x and denominator 34x2+13\sqrt{4x^{2}+1} separately:

ddx(4x)=4,ddx(34x2+1)=12x4x2+1\frac{d}{dx}\left(4x\right)=4,\qquad \frac{d}{dx}\left(3\sqrt{4x^{2}+1}\right)=\frac{12x}{\sqrt{4x^{2}+1}}

so the rule sends us to

lim⁡x→∞412x4x2+1=lim⁡x→∞44x2+112x=lim⁡x→∞4x2+13x\lim_{x\to\infty}\frac{4}{\dfrac{12x}{\sqrt{4x^{2}+1}}}=\lim_{x\to\infty}\frac{4\sqrt{4x^{2}+1}}{12x}=\lim_{x\to\infty}\frac{\sqrt{4x^{2}+1}}{3x}

which is the limit we started with (only the harmless +5+5 has gone). A third application returns to 4x34x2+1\frac{4x}{3\sqrt{4x^{2}+1}}, and so on forever. The rule loops: it is valid at each step but never terminates.

Step 4 — evaluate correctly by algebra. For x>0x>0 we have x=x2x=\sqrt{x^{2}}, so

4x2+1=x2(4+1x2)=x4+1x2\sqrt{4x^{2}+1}=\sqrt{x^{2}\left(4+\frac{1}{x^{2}}\right)}=x\sqrt{4+\frac{1}{x^{2}}}

(the positive root, since x>0x>0). Hence for x>0x>0

4x2+13x+5=x4+1x2x(3+5x)=4+1x23+5x\frac{\sqrt{4x^{2}+1}}{3x+5}=\frac{x\sqrt{4+\dfrac{1}{x^{2}}}}{x\left(3+\dfrac{5}{x}\right)}=\frac{\sqrt{4+\dfrac{1}{x^{2}}}}{3+\dfrac{5}{x}}

As x→∞x\to\infty, 1x2→0\frac{1}{x^{2}}\to 0 and 5x→0\frac{5}{x}\to 0; the denominator tends to 3≠03\ne 0, so the Quotient Law applies:

lim⁡x→∞4x2+13x+5=43=23\lim_{x\to\infty}\frac{\sqrt{4x^{2}+1}}{3x+5}=\frac{\sqrt{4}}{3}=\frac{2}{3}

Answer: 23\dfrac{2}{3}.

Independent check. At x=1000x=1000: 40000013005=2000.000253005≈0.66556\frac{\sqrt{4000001}}{3005}=\frac{2000.00025}{3005}\approx 0.66556, and at x=105x=10^{5} the value is ≈0.6666556\approx 0.6666556 — converging to 23=0.6667\frac{2}{3}=0.6667. Moral: looping is a signal to abandon the rule, not to keep differentiating. Warning for x→−∞x\to-\infty: there 4x2+1=∣x∣4+1x2=−x4+1x2\sqrt{4x^{2}+1}=|x|\sqrt{4+\frac{1}{x^{2}}}=-x\sqrt{4+\frac{1}{x^{2}}} and the limit is −23-\frac{2}{3}.

Consider lim⁡x→∞x+sin⁡xx−cos⁡x\displaystyle\lim_{x\to\infty}\frac{x+\sin x}{x-\cos x}. Show that l'Hopital's Rule fails to settle it — naming precisely which hypotheses break — and then find the limit by another method.

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Differentiate top and bottom and look hard at the resulting expression: ask both whether it has a limit and whether its denominator is ever zero far out.

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Step 1 — the form. Since −1≤sin⁡x≤1-1\le\sin x\le 1, we have x−1≤x+sin⁡x≤x+1x-1\le x+\sin x\le x+1, so x+sin⁡x→+∞x+\sin x\to+\infty; likewise x−cos⁡x→+∞x-\cos x\to+\infty. The form is ∞∞\frac{\infty}{\infty}: indeterminate, so the rule is worth considering.

Step 2 — what the rule produces. With f(x)=x+sin⁡xf(x)=x+\sin x and g(x)=x−cos⁡xg(x)=x-\cos x,

f′(x)=1+cos⁡x,g′(x)=1+sin⁡xf'(x)=1+\cos x,\qquad g'(x)=1+\sin x

so l'Hopital would point at lim⁡x→∞1+cos⁡x1+sin⁡x\displaystyle\lim_{x\to\infty}\frac{1+\cos x}{1+\sin x}.

Step 3 — hypothesis 2 fails. g′(x)=1+sin⁡x=0g'(x)=1+\sin x=0 whenever x=3π2+2kπx=\frac{3\pi}{2}+2k\pi for an integer kk. These points occur arbitrarily far out, so there is no interval (N,∞)\left(N,\infty\right) on which g′g' is nonvanishing. The theorem's second hypothesis is violated on every neighbourhood of ∞\infty.

Step 4 — hypothesis 4 fails too. Even ignoring the zeros, the quotient 1+cos⁡x1+sin⁡x\frac{1+\cos x}{1+\sin x} has no limit. Along x=2kπx=2k\pi it equals 1+11+0=2\frac{1+1}{1+0}=2; along x=(2k+1)πx=\left(2k+1\right)\pi it equals 1−11+0=0\frac{1-1}{1+0}=0. Two subsequences with different limits, so lim⁡x→∞f′g′\lim_{x\to\infty}\frac{f'}{g'} does not exist (and is not ±∞\pm\infty).

Step 5 — the correct conclusion about the rule. Because hypotheses 2 and 4 both fail, l'Hopital's Rule says nothing at all. In particular it is a logical error to conclude that the original limit does not exist.

Step 6 — evaluate by algebra and the Squeeze Theorem. For x>0x>0, divide numerator and denominator by xx:

x+sin⁡xx−cos⁡x=1+sin⁡xx1−cos⁡xx\frac{x+\sin x}{x-\cos x}=\frac{1+\dfrac{\sin x}{x}}{1-\dfrac{\cos x}{x}}

Since ∣sin⁡x∣≤1\left|\sin x\right|\le 1 we get −1x≤sin⁡xx≤1x-\frac{1}{x}\le\frac{\sin x}{x}\le\frac{1}{x} for x>0x>0, and both bounds tend to 00, so by the Squeeze Theorem sin⁡xx→0\frac{\sin x}{x}\to 0; identically cos⁡xx→0\frac{\cos x}{x}\to 0. The denominator therefore tends to 1≠01\ne 0, so the Quotient Law applies:

lim⁡x→∞x+sin⁡xx−cos⁡x=1+01−0=1\lim_{x\to\infty}\frac{x+\sin x}{x-\cos x}=\frac{1+0}{1-0}=1

Answer: the limit is 11; l'Hopital's Rule is simply inapplicable here.

Independent check. At x=1000x=1000: 1000+sin⁡10001000−cos⁡1000=1000.827999.438≈1.00139\frac{1000+\sin 1000}{1000-\cos 1000}=\frac{1000.827}{999.438}\approx 1.00139, and the deviation from 11 is at most about 2x\frac{2}{x}, which tends to 00.

Find all constants aa and bb for which

lim⁡x→0sin⁡2x+ax+bx3x3=0.\lim_{x\to 0}\frac{\sin 2x+ax+bx^{3}}{x^{3}}=0 .

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Apply the rule one stage at a time, and at each stage ask what the numerator must do for the quotient to stay indeterminate — a nonzero numerator over a vanishing denominator forces an infinite limit.

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Write f(x)=sin⁡2x+ax+bx3f(x)=\sin 2x+ax+bx^{3} and g(x)=x3g(x)=x^{3}. Both are differentiable everywhere, and g′(x)=3x2≠0g'(x)=3x^{2}\ne 0 for all x≠0x\ne 0, so hypotheses 1 and 2 hold on any interval around 00 for every choice of a,ba,b.

Stage 0. f(0)=sin⁡0+0+0=0f(0)=\sin 0+0+0=0 and g(0)=0g(0)=0, and both are continuous, so the form is 00\frac{0}{0} for every aa and bb.

Stage 1. f′(x)=2cos⁡2x+a+3bx2f'(x)=2\cos 2x+a+3bx^{2} and g′(x)=3x2g'(x)=3x^{2}, so the rule points at

lim⁡x→02cos⁡2x+a+3bx23x2\lim_{x\to 0}\frac{2\cos 2x+a+3bx^{2}}{3x^{2}}

The denominator →0+\to 0^{+} and the numerator →2+a\to 2+a.

Case 2+a≠02+a\ne 0. Then the form is nonzero0+\frac{\text{nonzero}}{0^{+}}, which is determinate and infinite: the quotient tends to +∞+\infty if 2+a>02+a>0 and to −∞-\infty if 2+a<02+a<0. Since hypothesis 4 permits L=±∞L=\pm\infty, l'Hopital's Rule then tells us the original limit is also ±∞\pm\infty — certainly not 00. So we must have

a=−2a=-2

Stage 2. With a=−2a=-2 the quotient is 2cos⁡2x−2+3bx23x2\dfrac{2\cos 2x-2+3bx^{2}}{3x^{2}}, whose numerator →2−2+0=0\to 2-2+0=0: the form is 00\frac{0}{0} again, and the new denominator derivative is 6x≠06x\ne 0 for x≠0x\ne 0. Differentiating:

lim⁡x→02cos⁡2x−2+3bx23x2=lim⁡x→0−4sin⁡2x+6bx6x\lim_{x\to 0}\frac{2\cos 2x-2+3bx^{2}}{3x^{2}}=\lim_{x\to 0}\frac{-4\sin 2x+6bx}{6x}

Stage 3 — re-check. Numerator →0\to 0 and denominator →0\to 0: still 00\frac{0}{0}, and the next denominator derivative is 6≠06\ne 0. Differentiating once more:

lim⁡x→0−4sin⁡2x+6bx6x=lim⁡x→0−8cos⁡2x+6b6=−8+6b6\lim_{x\to 0}\frac{-4\sin 2x+6bx}{6x}=\lim_{x\to 0}\frac{-8\cos 2x+6b}{6}=\frac{-8+6b}{6}

This last expression is continuous, so no indeterminacy remains and the chain of applications is valid from the bottom up. The original limit therefore equals 6b−86\dfrac{6b-8}{6}, and setting this to 00 gives

6b=8⇒b=436b=8\qquad\Rightarrow\qquad b=\frac{4}{3}

Answer: a=−2a=-2 and b=43b=\dfrac{4}{3}.

Independent check with the Maclaurin series. Since

sin⁡2x=2x−(2x)36+(2x)5120−⋯=2x−4x33+4x515−⋯\sin 2x=2x-\frac{\left(2x\right)^{3}}{6}+\frac{\left(2x\right)^{5}}{120}-\cdots=2x-\frac{4x^{3}}{3}+\frac{4x^{5}}{15}-\cdots

the numerator is (2+a)x+(b−43)x3+415x5−⋯\left(2+a\right)x+\left(b-\frac{4}{3}\right)x^{3}+\frac{4}{15}x^{5}-\cdots, so

sin⁡2x+ax+bx3x3=2+ax2+(b−43)+415x2−⋯\frac{\sin 2x+ax+bx^{3}}{x^{3}}=\frac{2+a}{x^{2}}+\left(b-\frac{4}{3}\right)+\frac{4}{15}x^{2}-\cdots

For the limit to exist and be 00 we need 2+a=02+a=0 and b−43=0b-\frac{4}{3}=0: the same answer. Numerically with a=−2a=-2, b=43b=\frac{4}{3} and x=0.1x=0.1, the numerator is 0.19866933−0.2+0.00133333=2.664×10−60.19866933-0.2+0.00133333=2.664\times 10^{-6}, and dividing by x3=10−3x^{3}=10^{-3} gives 0.0026640.002664 — matching the predicted 415x2=0.002667\frac{4}{15}x^{2}=0.002667, which indeed tends to 00.

Evaluate lim⁡x→0+(1x)tan⁡x\displaystyle\lim_{x\to 0^{+}}\left(\frac{1}{x}\right)^{\tan x}.

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Work out what the base and the exponent each tend to before touching anything. After taking logarithms you will meet a 0⋅∞0\cdot\infty product — turning tan⁡x\tan x into cot⁡x\cot x downstairs is the cleanest move.

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Step 1 — identify the form. As x→0+x\to 0^{+}: the base 1x→+∞\frac{1}{x}\to+\infty and the exponent tan⁡x→0\tan x\to 0. The form is ∞0\infty^{0}: indeterminate. (Do not write ∞0=1\infty^{0}=1; compare (1x)1/ln⁡(1/x)=e\left(\frac{1}{x}\right)^{1/\ln\left(1/x\right)}=e, which also has form ∞0\infty^{0} but equals ee.)

Step 2 — take logarithms. For 0<x<π20<x<\frac{\pi}{2} the base is positive and tan⁡x\tan x is defined. Let y=(1x)tan⁡xy=\left(\frac{1}{x}\right)^{\tan x}; then

ln⁡y=tan⁡x⋅ln⁡(1x)=−tan⁡xln⁡x\ln y=\tan x\cdot\ln\left(\frac{1}{x}\right)=-\tan x\ln x

with form 0⋅(+∞)0\cdot\left(+\infty\right): indeterminate, so make it a quotient.

Step 3 — make a quotient. Keep the logarithm upstairs and send tan⁡x\tan x downstairs as cot⁡x\cot x:

ln⁡y=−ln⁡xcot⁡x\ln y=\frac{-\ln x}{\cot x}

As x→0+x\to 0^{+}: −ln⁡x→+∞-\ln x\to+\infty and cot⁡x=cos⁡xsin⁡x→+∞\cot x=\frac{\cos x}{\sin x}\to+\infty, so this is the ∞∞\frac{\infty}{\infty} case. On (0,π2)\left(0,\frac{\pi}{2}\right) both functions are differentiable, and ddxcot⁡x=−csc⁡2x\frac{d}{dx}\cot x=-\csc^{2}x is never 00 there, so hypotheses 1 and 2 hold.

Step 4 — apply the rule.

\lim_{x\to 0^{+}}\frac{-\ln x}{\cot x} &= \lim_{x\to 0^{+}}\frac{-\dfrac{1}{x}}{-\csc^{2}x}\\ &= \lim_{x\to 0^{+}}\frac{1}{x}\cdot\sin^{2}x\\ &= \lim_{x\to 0^{+}}\frac{\sin x}{x}\cdot\sin x\\ &= 1\cdot 0=0 \end{aligned}$$ (The splitting in the last line is allowed because both factors have limits: $\frac{\sin x}{x}\to 1$ and $\sin x\to 0$.) So $\lim_{x\to 0^{+}}\ln y=0$. **Step 5 — undo the logarithm.** Since $\exp$ is continuous, $$\lim_{x\to 0^{+}}\left(\frac{1}{x}\right)^{\tan x}=e^{0}=1$$ **Answer: $1$.** *Independent check.* Write $\ln y=-\tan x\ln x=-\dfrac{\tan x}{x}\cdot x\ln x$. Here $\frac{\tan x}{x}\to 1$ and $x\ln x\to 0$ (a standard limit), so $\ln y\to-1\cdot 0=0$ and $y\to 1$, as found. Numerically at $x=0.001$: $\tan\left(0.001\right)\ln\left(1000\right)\approx 0.0069078$, so $y\approx e^{0.0069}\approx 1.0069$ — approaching $1$ from above.