L'Hôpital's Rule and Indeterminate Forms

Quick answer

L'Hôpital's rule says that if f(x)/g(x) takes the form 0/0 or ∞/∞ as x → a, then its limit equals the limit of f′(x)/g′(x), when that limit exists. The reason is local linearity: near a, f and g behave like their tangent lines, so their ratio behaves like the ratio of their slopes. Other indeterminate forms are first rewritten as quotients — a product as a fraction, a difference over a common denominator, a power through logarithms.

What you'll learn

Check that a limit has the form 0/0 or ∞/∞ before using L'Hôpital's rule
Apply the rule, repeatedly if needed
Rewrite 0·∞ and ∞ − ∞ forms as quotients
Use logarithms for the forms 1^∞, 0⁰ and ∞⁰

The rule

Some limits resist algebra: $\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x}$ is $\tfrac{0}{0}$ , and nothing factors. Derivatives settle it.

L'Hôpital's rule

Suppose $f$ and $g$ are differentiable near $a$ (except possibly at $a$ ), with $g'(x) \ne 0$ there, and $\tfrac{f(x)}{g(x)}$ has the form $\tfrac{0}{0}$ or $\tfrac{\pm\infty}{\pm\infty}$ as $x \to a$ . Then

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

provided the limit on the right exists or is $\pm\infty$ . The same holds for one-sided limits and for $x \to \pm\infty$ .

So $\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = 1$ , confirming the estimate from the first lesson.

Why the slopes decide it

Take the simplest case: $f(a) = g(a) = 0$ , and both are differentiable at $a$ with $g'(a) \ne 0$ . Near $a$ , each function is close to its tangent line, and both tangent lines pass through $(a, 0)$ :

f(x) \approx f'(a)(x - a), \qquad g(x) \approx g'(a)(x - a)

The factor $x - a$ cancels in the ratio, leaving $\tfrac{f'(a)}{g'(a)}$ . Exactly,

\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{\tfrac{f(x) - f(a)}{x - a}}{\tfrac{g(x) - g(a)}{x - a}} \;\longrightarrow\; \frac{f'(a)}{g'(a)}

When both functions vanish at the same point, their ratio is decided by how fast each one leaves zero — their derivatives. The general statement, for $\tfrac{\infty}{\infty}$ and when $f'(a)$ does not exist, is proved with Cauchy’s mean value theorem.

Other indeterminate forms

The rule handles only quotients. The remaining indeterminate forms are rewritten into one first:

Form	Rewrite
$0 \cdot \infty$	$f g = \tfrac{f}{1/g}$ , a $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ form
$\infty - \infty$	combine over a common denominator, or factor
$1^\infty$ , $0^0$ , $\infty^0$	take $\ln$ : $\ln(f^g) = g \ln f$ , a $0 \cdot \infty$ form

For the power forms, find the limit $L$ of $\ln y$ , and then the original limit is $e^L$ .

Worked examples

Example 3 — The limit that defines e

Find $\displaystyle\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$ .

The form is $1^\infty$ . Let $y = \left(1 + \tfrac{1}{x}\right)^x$ , so

\ln y = x \ln\!\left(1 + \frac{1}{x}\right) = \frac{\ln\left(1 + \frac{1}{x}\right)}{1/x}

a $\tfrac{0}{0}$ form. Apply the rule, simplifying the chain-rule factor $-\tfrac{1}{x^2}$ that appears on top and bottom:

\lim_{x \to \infty} \frac{\frac{-1/x^2}{1 + 1/x}}{-1/x^2} = \lim_{x \to \infty} \frac{1}{1 + 1/x} = 1

So $\ln y \to 1$ and $y \to e^1 = e$ .

Common mistakes

Practice problems

Find $\lim_{x \to 0} \tfrac{\sin 4x}{x}$ .

Answer

$4$

Full solution

Form $\tfrac{0}{0}$ . The rule gives $\tfrac{4\cos 4x}{1} \to 4$ .
Find $\lim_{x \to 2} \tfrac{x^3 - 8}{x - 2}$ .

Answer

$12$

Full solution

Form $\tfrac{0}{0}$ . The rule gives $\tfrac{3x^2}{1} \to 12$ . (Factoring gives $x^2 + 2x + 4 \to 12$ too.)
Find $\lim_{x \to \infty} \tfrac{\ln x}{\sqrt{x}}$ .

Answer

$0$

Full solution

Form $\tfrac{\infty}{\infty}$ . The rule gives $\tfrac{1/x}{1/(2\sqrt{x})} = \tfrac{2\sqrt{x}}{x} = \tfrac{2}{\sqrt{x}} \to 0$ .
Find $\lim_{x \to 0} \tfrac{1 - \cos x}{x^2}$ .

Answer

$\tfrac{1}{2}$

Full solution

Form $\tfrac{0}{0}$ . Once: $\tfrac{\sin x}{2x}$ , still $\tfrac{0}{0}$ . Twice: $\tfrac{\cos x}{2} \to \tfrac{1}{2}$ .
Find $\lim_{x \to \infty} x e^{-x}$ .

Answer

$0$

Full solution

Form $\infty \cdot 0$ . Rewrite as $\tfrac{x}{e^x}$ , form $\tfrac{\infty}{\infty}$ . The rule gives $\tfrac{1}{e^x} \to 0$ .
Find $\lim_{x \to 0^+} x^x$ .

Answer

$1$

Full solution

Form $0^0$ . $\ln(x^x) = x\ln x \to 0$ (Example 2). So $x^x \to e^0 = 1$ .
Find $\lim_{x \to 0} \left(\tfrac{1}{x} - \tfrac{1}{\sin x}\right)$ .

Answer

$0$

Full solution

Form $\infty - \infty$ . Combine: $\tfrac{\sin x - x}{x \sin x}$ , form $\tfrac{0}{0}$ .

Once: $\tfrac{\cos x - 1}{\sin x + x\cos x}$ , still $\tfrac{0}{0}$ . Twice: $\tfrac{-\sin x}{2\cos x - x\sin x} \to \tfrac{0}{2} = 0$ .
Find $\lim_{x \to \infty} \left(1 + \tfrac{3}{x}\right)^x$ .

Answer

$e^3$

Full solution

$\ln y = x\ln\!\left(1 + \tfrac{3}{x}\right) = \tfrac{\ln(1 + 3/x)}{1/x}$ , form $\tfrac{0}{0}$ . The rule gives $\tfrac{3}{1 + 3/x} \to 3$ , so $y \to e^3$ .
Find $\lim_{x \to \infty} \tfrac{3x^2 + x}{5x^2 - 2}$ with L’Hôpital’s rule, and check it against the leading-coefficient method.

Answer

$\tfrac{3}{5}$

Full solution

Once: $\tfrac{6x + 1}{10x}$ , still $\tfrac{\infty}{\infty}$ . Twice: $\tfrac{6}{10} = \tfrac{3}{5}$ . The ratio of leading coefficients is also $\tfrac{3}{5}$ .
A student computes $\lim_{x \to 0} \tfrac{\cos x}{x + 1}$ with L’Hôpital’s rule and gets $\tfrac{-\sin 0}{1} = 0$ . What went wrong?

Hint

What form does the limit have?

Answer

The form is $\tfrac{1}{1}$ , not indeterminate. The limit is $1$ .

Full solution

Substitution gives $\tfrac{\cos 0}{0 + 1} = 1$ , a determinate value.

The rule applies only to $\tfrac{0}{0}$ and $\tfrac{\infty}{\infty}$ ; used here it produces a wrong answer.

Frequently asked questions

What is L'Hôpital's rule?

If f(x)/g(x) has the form 0/0 or ∞/∞ as x approaches a, then lim f(x)/g(x) = lim f′(x)/g′(x), provided the second limit exists or is infinite.

When can I use L'Hôpital's rule?

Only for the indeterminate forms 0/0 and ∞/∞. For any other form, rewrite first; using it on a determinate form gives wrong answers.

Do I use the quotient rule in L'Hôpital's rule?

No. Differentiate the numerator and the denominator separately; the fraction is not differentiated as a quotient.

How do I handle 1^∞ or 0⁰?

Take the natural log of the expression, which turns the power into a product. Find that limit, then exponentiate the answer.

Can I apply the rule more than once?

Yes, as long as each new quotient is still 0/0 or ∞/∞.

What to learn next

The Mean Value Theorem and the Extreme Value TheoremRolle's theorem, the Mean Value Theorem and the Extreme Value Theorem: what each guarantees, why the hypotheses matter, and what follows from them.

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