When can you use L'Hôpital's rule?

Only when substituting gives frac00 or (pm∞)/(pm∞). Other undecided forms must be rewritten as one of these first.

Is L'Hôpital's rule the same as the quotient rule?

No. L'Hôpital's rule takes the derivative of the top and the derivative of the bottom separately, then divides. The quotient rule finds the derivative of the whole fraction.

Can you use L'Hôpital's rule more than once?

Yes, as long as each new limit is still frac00 or frac∞∞. Check the form before every application.

What if L'Hôpital's rule keeps going in circles?

Some limits, like fracsqrtx^2+1x as x→∞, flip back and forth forever. Then use algebra instead, such as dividing by the highest power.

Can I use L'Hôpital's rule on limits like sin(x)/x?

It gives the right answer, 1, but many courses don't allow it there, because the derivative of sin x is itself proved using that limit. Check what your class expects.

L'Hôpital's Rule: How It Works and When to Use It

What problem does L'Hôpital's rule solve?

When substituting into a limit gives $\frac{0}{0}$ , the usual move is algebra: factor and cancel, or use a conjugate (see how to evaluate limits). But some limits, such as $lim_{x \to 0} \frac{e^{2 x} - 1}{\sin x}$ , mix functions that no algebra can simplify together. L'Hôpital's rule (pronounced lo-pee-TAHL) handles these using derivatives.

What is an indeterminate form?

Forms like $\frac{0}{0}$ and $\frac{\infty}{\infty}$ are called indeterminate, because knowing the form doesn't determine the answer. In $\frac{0}{0}$ , the top pulls the fraction toward $0$ while the bottom pulls it toward something huge, and the limit depends on which shrinks faster. For example, as $x \to 0$ , all three of $\frac{x^{2}}{x}$ , $\frac{x}{x}$ and $\frac{x}{x^{2}}$ start as $\frac{0}{0}$ , yet their limits are $0$ , $1$ , and nonexistent.

What does L'Hôpital's rule say?

L'Hôpital's rule

Suppose

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

has the form

\frac{0}{0}

\frac{\pm \infty}{\pm \infty}

, and

f

and

g

are differentiable near

a

with

g^{'} (x) \neq 0

there. Then

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

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

, provided the right side exists (or is

\pm \infty

). The same holds for one-sided limits and for

x \to \pm \infty

Why it works

Take the $\frac{0}{0}$ case with $f (a) = g (a) = 0$ and $g^{'} (a) \neq 0$ .
Near $a$ , each function is close to its tangent line: $f (x) \approx f (a) + f^{'} (a) (x - a) =$ $f^{'} (a) (x - a)$ , and likewise $g (x) \approx g^{'} (a) (x - a)$ .
So $\frac{f (x)}{g (x)} \approx \frac{f^{'} (a) (x - a)}{g^{'} (a) (x - a)} =$ $\frac{f^{'} (a)}{g^{'} (a)}$ , and the approximation gets better as $x \to a$ .
In words: when both functions are heading to $0$ , what matters is how fast each one is heading there, which is exactly what their derivatives measure.

How do you use L'Hôpital's rule, step by step?

Check the form by substituting: it must be $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
Differentiate the top and the bottom separately.
Find the new limit, usually by substituting.
If the new limit is still $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , check again and repeat.

Worked example: Exponential over trig

Find $lim_{x \to 0} \frac{e^{2 x} - 1}{\sin x}$ .

Check: $\frac{e^{0} - 1}{\sin 0} =$ $\frac{1 - 1}{0} =$ $\frac{0}{0}$ . ✓
Derivatives: top $\frac{d}{d x} (e^{2 x} - 1) =$ $2 e^{2 x}$ (chain rule); bottom $\frac{d}{d x} \sin x =$ $\cos x$ .
New limit: $lim_{x \to 0} \frac{2 e^{2 x}}{\cos x} =$ $\frac{2 \cdot 1}{1} =$ $2$ .

Answer: $2$

Worked example: Infinity over infinity

Find $lim_{x \to \infty} \frac{\ln x}{x}$ .

Check: as $x \to \infty$ , both $\ln x$ and $x$ grow without bound: $\frac{\infty}{\infty}$ . ✓
Derivatives: top $\frac{1}{x}$ ; bottom $1$ .
New limit: $lim_{x \to \infty} \frac{1 / x}{1} =$ $lim_{x \to \infty} \frac{1}{x} =$ $0$ .

Answer: $0$ : $x$ grows much faster than $\ln x$ .

Worked example: Using the rule twice

Find $lim_{x \to 0} \frac{1 - \cos x}{x^{2}}$ .

Check: $\frac{1 - 1}{0} =$ $\frac{0}{0}$ . ✓
Derivatives: top $\sin x$ ; bottom $2 x$ . New limit $lim_{x \to 0} \frac{\sin x}{2 x}$ , which is again $\frac{0}{0}$ .
Apply the rule again: top $\cos x$ ; bottom $2$ .
New limit: $\frac{\cos 0}{2} =$ $\frac{1}{2}$ .

Answer: $\frac{1}{2}$

Common mistake

Using the quotient rule. L'Hôpital's rule differentiates the top and the bottom separately: $\frac{f^{'}}{g^{'}}$ , not ${(\frac{f}{g})}^{'}$ . Using the quotient rule gives a completely different (and wrong) expression.

Why do you have to check the form first?

Because the rule is only true for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ . Used anywhere else, it gives wrong answers.

Worked example: When not to use it

Find $lim_{x \to 0} \frac{x + 1}{x + 2}$ .

Substitute: $\frac{0 + 1}{0 + 2} =$ $\frac{1}{2}$ . That's an ordinary number, so the limit is $\frac{1}{2}$ .
Wrongly applying L'Hôpital's rule would give $\frac{1}{1} = 1$ , which is incorrect, because the form wasn't $\frac{0}{0}$ .

Answer: $\frac{1}{2}$

Which grows fastest: powers, exponentials or logarithms?

L'Hôpital's rule settles the long-run race between functions. Applying it to $\frac{x^{2}}{e^{x}}$ twice gives $\frac{2 x}{e^{x}}$ and then $\frac{2}{e^{x}} \to 0$ : the exponential wins. The same happens for any power, and logarithms lose to every positive power. As $x \to \infty$ :

\ln x ≪ x^{p} ≪ e^{x}

for any power $p > 0$ , where $≪$ means "grows much more slowly than", so the slower one divided by the faster one goes to $0$ .

The ratio

y = \frac{x^{2}}{e^{x}}

rises briefly, peaks at

x = 2

, then falls toward

0

e^{x}

eventually outgrows

x^{2}

completely.

How do you handle other indeterminate forms?

Other undecided forms, such as $0 \cdot \infty$ , $\infty - \infty$ , $1^{\infty}$ , $0^{0}$ and $\infty^{0}$ , can't use the rule directly. Rewrite them as a fraction first:

Form	Rewrite	Example
$0 \cdot \infty$	move one factor to the bottom: $f g = \frac{f}{1 / g}$	$x \ln x =$ $\frac{\ln x}{1 / x}$
$\infty - \infty$	combine into one fraction (common bottom or conjugate)	$\frac{1}{x} - \frac{1}{\sin x} =$ $\frac{\sin x - x}{x \sin x}$
$1^{\infty}$ , $0^{0}$ , $\infty^{0}$	take $\ln$ : the exponent comes down as a product	$\ln ((1 + \frac{1}{x})^{x}) =$ $x \ln (1 + \frac{1}{x})$

Worked example: Zero times infinity

Find $lim_{x \to 0^{+}} x \ln x$ .

As $x \to 0^{+}$ , $x \to 0$ and $\ln x \to - \infty$ : the form $0 \cdot \infty$ .
Rewrite as a fraction: $x \ln x =$ $\frac{\ln x}{1 / x}$ , which has the form $\frac{- \infty}{\infty}$ .
Derivatives: top $\frac{1}{x}$ ; bottom $- \frac{1}{x^{2}}$ .
Simplify: $\frac{1 / x}{- 1 / x^{2}} =$ $\frac{1}{x} \cdot (- x^{2}) =$ $- x$ , and $- x \to 0$ .

Answer: $0$

Tip

L'Hôpital's rule isn't always the quickest route. For $\frac{x^{2} - 9}{x - 3}$ , factoring is faster; for $\frac{3 x^{2} + 1}{5 x^{2} - x}$ as $x \to \infty$ , comparing degrees (limits at infinity) gives $\frac{3}{5}$ at a glance.

L'Hôpital's rule: how it works and when to use it