Unit 4, lesson 43

Indeterminate Forms and l’Hôpital’s Rule

When a limit has a form like 0 over 0 that does not decide its value, how l’Hôpital’s rule settles it, which hypotheses it needs, and how to handle products, differences and powers.

  • Theory part 85
  • Practice part 86
  • Limits and Approximation

As x→0x \to 0 both sin⁡x\sin x and xx shrink to 00, so their quotient is a race between two vanishing quantities. Knowing that both go to 00 says nothing about who wins. The derivatives measure how fast each one is shrinking, and l’Hôpital’s rule turns that idea into a precise tool for limits that algebra alone cannot open.

Key terms

The forms 0 over 0 and infinity over infinity

Definition Let ff and gg be defined near aa, except possibly at aa, with g(x)≠0g(x) \neq 0 there. The limit

lim⁡x→af(x)g(x)\lim_{x \to a} \frac{f(x)}{g(x)}

is of indeterminate type 00\frac{0}{0} if f(x)→0f(x) \to 0 and g(x)→0g(x) \to 0 as x→ax \to a. It is of type ∞∞\frac{\infty}{\infty} if ∣f(x)∣→∞\lvert f(x) \rvert \to \infty and ∣g(x)∣→∞\lvert g(x) \rvert \to \infty as x→ax \to a. Indeterminate means that the type alone does not determine the limit, which may be any real number, ±∞\pm\infty, or may not exist.

In plain words Zero over zero is a question, not an answer. Different functions of this type give different limits.

Example All three of these limits are of type 00\frac{0}{0} at 00.

lim⁡x→0xx=1,lim⁡x→0x2x=0,lim⁡x→0+xx2=∞.\lim_{x \to 0} \frac{x}{x} = 1, \qquad \lim_{x \to 0} \frac{x^2}{x} = 0, \qquad \lim_{x \to 0^+} \frac{x}{x^2} = \infty.

The rule and its hypotheses

Definition This is l’Hôpital’s rule. Let II be an open interval containing aa. Suppose ff and gg are differentiable on II except possibly at aa, and g′(x)≠0g'(x) \neq 0 for every xx in II with x≠ax \neq a. Suppose the limit of f(x)/g(x)f(x)/g(x) as x→ax \to a is of type 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. If

lim⁡x→af′(x)g′(x)=L,\lim_{x \to a} \frac{f'(x)}{g'(x)} = L,

where LL is a real number, ∞\infty or −∞-\infty, then also

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

In the ∞∞\frac{\infty}{\infty} case it is in fact enough that ∣g(x)∣→∞\lvert g(x) \rvert \to \infty, whatever ff does.

In plain words For a 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} quotient, you may replace top and bottom by their derivatives (separately, not with the quotient rule), as long as the new limit exists or is infinite and the bottom derivative is not zero near aa.

Example The limit of (ex−1)/x(e^x - 1)/x as x→0x \to 0 is of type 00\frac{0}{0}, both functions are differentiable everywhere, and g′(x)=1≠0g'(x) = 1 \neq 0. So

lim⁡x→0ex−1x=lim⁡x→0ex1=1.\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = 1.

The type check matters. The limit

lim⁡x→0x+1x+2=12\lim_{x \to 0} \frac{x + 1}{x + 2} = \frac{1}{2}

is not indeterminate, and differentiating anyway would give the wrong value 11.

One-sided limits and limits at infinity

Definition The rule holds with the same conclusion for one-sided limits x→a+x \to a^+ and x→a−x \to a^-, using an interval (a,a+δ)(a, a + \delta) or (a−δ,a)(a - \delta, a) in place of II, and for x→∞x \to \infty or x→−∞x \to -\infty, using an interval (N,∞)(N, \infty) or (−∞,N)(-\infty, N). The limit LL may be infinite in every case.

In plain words The rule works from one side and far out on the number line too, and it can show that a limit is infinite.

Example As x→∞x \to \infty the quotient (ln⁡x)/x(\ln x)/\sqrt{x} is of type ∞∞\frac{\infty}{\infty}, and g′(x)=1/(2x)≠0g'(x) = 1/(2\sqrt{x}) \neq 0 for x>0x > 0. So

lim⁡x→∞ln⁡xx=lim⁡x→∞1/x1/(2x)=lim⁡x→∞2x=0.\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} = \lim_{x \to \infty} \frac{1/x}{1/(2\sqrt{x})} = \lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0.

From the right at 00 the quotient sin⁡x/(1−cos⁡x)\sin x/(1 - \cos x) is of type 00\frac{0}{0}, and g′(x)=sin⁡x≠0g'(x) = \sin x \neq 0 on (0,π)(0, \pi). The quotient of derivatives satisfies

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

so the original limit is ∞\infty too.

Indeterminate products

Definition If lim⁡x→af(x)=0\lim_{x \to a} f(x) = 0 and lim⁡x→a∣g(x)∣=∞\lim_{x \to a} \lvert g(x) \rvert = \infty, the limit of f(x)g(x)f(x)g(x) is of indeterminate type 0⋅∞0 \cdot \infty. Rewrite it as a quotient,

f(x)g(x)=f(x)1/g(x)orf(x)g(x)=g(x)1/f(x),f(x)g(x) = \frac{f(x)}{1/g(x)} \quad \text{or} \quad f(x)g(x) = \frac{g(x)}{1/f(x)},

which is of type 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. The first form needs g(x)≠0g(x) \neq 0 near aa, which holds because ∣g(x)∣→∞\lvert g(x) \rvert \to \infty. The second needs f(x)≠0f(x) \neq 0 for xx near aa with x≠ax \neq a.

In plain words Move one factor into the denominator as its reciprocal. Choose the one whose derivative stays simple.

Example For the limit of xln⁡xx \ln x as x→0+x \to 0^+, write the product as (ln⁡x)/(1/x)(\ln x)/(1/x), of type ∞∞\frac{\infty}{\infty}. Then

lim⁡x→0+ln⁡x1/x=lim⁡x→0+1/x−1/x2=lim⁡x→0+(−x)=0.\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0.

Indeterminate differences

Definition If lim⁡x→af(x)=∞\lim_{x \to a} f(x) = \infty and lim⁡x→ag(x)=∞\lim_{x \to a} g(x) = \infty, the limit of f(x)−g(x)f(x) - g(x) is of indeterminate type ∞−∞\infty - \infty. Rewrite the difference as a single quotient, by a common denominator, by rationalizing, or by factoring, and then evaluate that quotient.

In plain words Two huge quantities can cancel in any proportion. Combine them into one fraction first.

Example Consider the limit

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

where both terms grow without bound. A common denominator gives

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

which is of type 00\frac{0}{0}. Its value is found in the section on repeated use below.

Indeterminate powers

Definition A limit of f(x)g(x)f(x)^{g(x)} with f(x)>0f(x) > 0 near aa is indeterminate when it is of type 000^0 (f→0+f \to 0^+, g→0g \to 0), ∞0\infty^0 (f→∞f \to \infty, g→0g \to 0) or 1∞1^\infty (f→1f \to 1, ∣g∣→∞\lvert g \rvert \to \infty). Write

f(x)g(x)=eg(x)ln⁡f(x).f(x)^{g(x)} = e^{g(x)\ln f(x)}.

Because the exponential function is continuous, if g(x)ln⁡f(x)→Mg(x)\ln f(x) \to M for a real number MM, then

lim⁡x→af(x)g(x)=eM.\lim_{x \to a} f(x)^{g(x)} = e^M.

If instead g(x)ln⁡f(x)→∞g(x)\ln f(x) \to \infty, the power tends to ∞\infty, and if it tends to −∞-\infty, the power tends to 00.

In plain words Take the logarithm to bring the exponent down, find that limit, then exponentiate the answer.

Example The limit lim⁡x→0+xx\lim_{x \to 0^+} x^x is of type 000^0. Here g(x)ln⁡f(x)=xln⁡x→0g(x)\ln f(x) = x\ln x \to 0 by the product example above, so

lim⁡x→0+xx=e0=1.\lim_{x \to 0^+} x^x = e^0 = 1.

Using the rule more than once

Definition If f′/g′f'/g' is again of type 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} and f′f' and g′g' satisfy the hypotheses, the rule may be applied to f′/g′f'/g', and so on. The hypotheses must be checked again at each step, and you must stop as soon as the quotient is no longer indeterminate.

In plain words Keep going while the form stays 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. Check the form every time.

Example Applied twice,

lim⁡x→01−cos⁡xx2=lim⁡x→0sin⁡x2x=lim⁡x→0cos⁡x2=12.\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{\sin x}{2x} = \lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}.

The difference example also needs two steps. Here the interval (0,π/4)(0, \pi/4) plays the role of (a,a+δ)(a, a + \delta). On it the two denominators that the rule differentiates, xsin⁡xx \sin x and then sin⁡x+xcos⁡x\sin x + x\cos x, have derivatives sin⁡x+xcos⁡x>0\sin x + x\cos x > 0 and 2cos⁡x−xsin⁡x>02\cos x - x\sin x > 0, so the hypothesis on g′g' holds at both steps, and

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

When the rule is inconclusive

Definition If f′(x)/g′(x)f'(x)/g'(x) has no limit as x→ax \to a, finite or infinite, then l’Hôpital’s rule says nothing about the limit of f(x)/g(x)f(x)/g(x), which may still exist. The rule can also go in circles, with each step giving back an equivalent form. In both cases use another method, such as algebra, bounds or the squeeze theorem.

In plain words A failed rule does not mean the limit fails to exist. It just means you need a different approach.

Example Take the quotient (x+sin⁡x)/x(x + \sin x)/x as x→∞x \to \infty. The quotient of derivatives is 1+cos⁡x1 + \cos x, which keeps swinging between 00 and 22 and has no limit. Yet for x>0x > 0

x+sin⁡xx=1+sin⁡xx→1,because∣sin⁡xx∣≤1x.\frac{x + \sin x}{x} = 1 + \frac{\sin x}{x} \to 1, \qquad \text{because} \quad \left\lvert \frac{\sin x}{x} \right\rvert \le \frac{1}{x}.

Similarly, for x2+1/x\sqrt{x^2 + 1}/x the rule gives x/x2+1x/\sqrt{x^2 + 1}, and applying it again gives back the original. Algebra settles it at once, since for x>0x > 0

x2+1x=1+1x2→1as x→∞.\frac{\sqrt{x^2 + 1}}{x} = \sqrt{1 + \frac{1}{x^2}} \to 1 \quad \text{as } x \to \infty.

Check yourself

  1. Find lim⁡x→0sin⁡3xx\lim_{x \to 0} \frac{\sin 3x}{x}.
Show the answer

It is of type 00\frac{0}{0} with g′(x)=1g'(x) = 1, so

lim⁡x→0sin⁡3xx=lim⁡x→03cos⁡3x1=3.\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{3\cos 3x}{1} = 3.
  1. Find lim⁡x→∞x2e−x\lim_{x \to \infty} x^2 e^{-x}.
Show the answer

Write it as x2/exx^2/e^x, of type ∞∞\frac{\infty}{\infty}, where the denominator has derivative ex≠0e^x \neq 0. Two applications give

lim⁡x→∞x2ex=lim⁡x→∞2xex=lim⁡x→∞2ex=0.\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x} = 0.
  1. Find lim⁡x→0+(1+2x)1/x\lim_{x \to 0^+} (1 + 2x)^{1/x}.
Show the answer

It is of type 1∞1^\infty. The exponent of ee is ln⁡(1+2x)/x\ln(1 + 2x)/x, of type 00\frac{0}{0}, and

lim⁡x→0+ln⁡(1+2x)x=lim⁡x→0+2/(1+2x)1=2.\lim_{x \to 0^+} \frac{\ln(1 + 2x)}{x} = \lim_{x \to 0^+} \frac{2/(1 + 2x)}{1} = 2.

So the limit is e2e^2.

How this lesson connects

It also uses theorem hypotheses.