L'Hospital's Rule

Some limits, like $\displaystyle\lim_{x \to 0} \frac{\sin 3x}{x}$ or $\displaystyle\lim_{x \to \infty} \frac{x^2}{e^x}$, can’t be found by substituting, and factoring doesn’t help. L’Hospital’s Rule (also spelled L’Hôpital’s, after the French mathematician; AP uses “L’Hospital”) turns these into easier limits using derivatives. It’s quick and powerful, but only when you check the form first.

Key ideas

Indeterminate forms

If substituting gives $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, the limit is indeterminate: the form alone doesn’t tell you the answer. It could be any number, or infinite. You saw how to handle some of these with algebraic techniques like factoring. L’Hospital’s Rule handles many more.

The rule

Suppose $f$ and $g$ are differentiable near $a$ (except possibly at $a$), $g'(x) \ne 0$ near $a$, and

$$\lim_{x \to a} f(x) = 0 \ \text{and}\ \lim_{x \to a} g(x) = 0, \qquad \text{or} \qquad \lim_{x \to a} f(x) = \pm\infty \ \text{and}\ \lim_{x \to a} g(x) = \pm\infty$$

Then

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

as long as the limit on the right exists (or is $\pm\infty$). The same rule works for $x \to \infty$ and for one-sided limits.

Notice: you differentiate the top and the bottom separately. This is not the quotient rule.

Why it works

For the $\dfrac{0}{0}$ case, think about local linearity. If $f(a) = g(a) = 0$, then near $a$ each function is close to its tangent line: $f(x) \approx f'(a)(x - a)$ and $g(x) \approx g'(a)(x - a)$. The $(x - a)$ factors cancel, leaving $\dfrac{f'(a)}{g'(a)}$.

Near $x = 1$, $\ln x \approx 1(x - 1)$ and $x^2 - 1 \approx 2(x - 1)$, so $\displaystyle\lim_{x \to 1} \frac{\ln x}{x^2 - 1} = \frac{1}{2}$.

How to use it

1. Check the form by substituting. Only continue if you get $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$.
2. Differentiate the numerator and the denominator separately.
3. Find the new limit. If it is still $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, you may apply the rule again (check the form each time).

On a test, show that the form is indeterminate before using the rule, for example: ”$\displaystyle\lim_{x \to 0}(e^x - 1) = 0$ and $\displaystyle\lim_{x \to 0} x = 0$, so L’Hospital’s Rule applies.” AP questions in particular often give values of $f$, $g$, $f'$, $g'$ in a table instead of formulas.

When not to use it

• When the form is not indeterminate. If substitution gives a number like $\dfrac{1}{1}$ or $\dfrac{5}{0}$, the rule doesn’t apply and will give a wrong answer.
• When an easier method works. Factoring or dividing by the highest power is often quicker.
• When the derivatives get messier each time and never resolve. Try rewriting the expression instead.

All trig in calculus uses radians. For example, $\dfrac{d}{dx}\sin x = \cos x$ is only true in radians.

Worked examples

Example 1: A trig limit

Find $\displaystyle\lim_{x \to 0} \frac{\sin 3x}{x}$.

Solution. Check the form: $\sin 0 = 0$ and the bottom is $0$, so the form is $\dfrac{0}{0}$. Apply the rule:

$$\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{3\cos 3x}{1} = 3\cos 0 = 3$$

Example 2: Applying the rule twice

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

Solution. Substituting gives $\dfrac{1 - 1 - 0}{0} = \dfrac{0}{0}$. Apply the rule:

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

This is still $\dfrac{0}{0}$ (since $e^0 - 1 = 0$), so apply it again:

$$\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$$

Example 3: An infinity over infinity form

Find $\displaystyle\lim_{x \to \infty} \frac{x^2}{e^x}$.

Solution. As $x \to \infty$, both $x^2$ and $e^x$ go to $\infty$: the form is $\dfrac{\infty}{\infty}$.

$$\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x} = 0$$

(The middle limit is still $\dfrac{\infty}{\infty}$, so we applied the rule a second time.) This shows that $e^x$ grows faster than $x^2$; in fact it grows faster than any power of $x$.

Example 4: When the rule doesn’t apply

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

Solution. Substitute: $\dfrac{\cos 0}{0 + 1} = \dfrac{1}{1} = 1$. That’s a number, not an indeterminate form, so the limit is $1$.

If you had wrongly used L’Hospital’s Rule, you’d get $\dfrac{-\sin 0}{1} = 0$, which is wrong. Always check the form first.

Common mistakes

Using the rule without checking the form. Example 4 shows how this gives a wrong answer. Substitute first, every time, including before a second application.

Using the quotient rule. L’Hospital’s Rule takes $\dfrac{f'(x)}{g'(x)}$, the derivative of the top over the derivative of the bottom, not the derivative of $\dfrac{f(x)}{g(x)}$.

Not showing the form in a written solution. Markers (including AP graders) expect you to state that both the numerator and denominator have limit $0$ (or both are infinite) before using the rule.

Applying it to forms like $\dfrac{5}{0}$. That form means the limit is infinite or doesn’t exist (check the signs on each side); it’s not a case for L’Hospital’s Rule.

Forgetting the chain rule when differentiating. $\dfrac{d}{dx}\sin 3x = 3\cos 3x$ and $\dfrac{d}{dx}e^{2x} = 2e^{2x}$.

Practice

1. (Warm-up) Which of these limits are indeterminate forms? Evaluate the one that is not.

• (a) $\displaystyle\lim_{x \to 1} \frac{x^2 - 1}{x + 1}$
• (b) $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$

Solution

(a) Substituting gives $\dfrac{0}{2} = 0$. Not indeterminate: the limit is $0$.

(b) Substituting gives $\dfrac{1 - 1}{0} = \dfrac{0}{0}$. Indeterminate (see question 4).

2. (Warm-up) Find $\displaystyle\lim_{x \to 0} \frac{\tan x}{x}$.

Solution

The form is $\dfrac{0}{0}$.

$$\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sec^2 x}{1} = \sec^2 0 = 1$$

3. (Core) Find $\displaystyle\lim_{x \to 1} \frac{\ln x}{x - 1}$.

Solution

$\ln 1 = 0$ and $1 - 1 = 0$, so the form is $\dfrac{0}{0}$.

$$\lim_{x \to 1} \frac{\ln x}{x - 1} = \lim_{x \to 1} \frac{1/x}{1} = 1$$

4. (Core) Find $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.

Solution

The form is $\dfrac{0}{0}$. Apply the rule:

$$\lim_{x \to 0} \frac{\sin x}{2x}$$

Still $\dfrac{0}{0}$, so apply it again:

$$\lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}$$

5. (Core) Find $\displaystyle\lim_{x \to \infty} \frac{3x^2 + 5}{2x^2 - x}$ two ways: with L’Hospital’s Rule, and by dividing by $x^2$.

Solution

The form is $\dfrac{\infty}{\infty}$.

L’Hospital’s Rule (twice):

$$\lim_{x \to \infty} \frac{6x}{4x - 1} = \lim_{x \to \infty} \frac{6}{4} = \frac{3}{2}$$

Dividing by $x^2$:

$$\lim_{x \to \infty} \frac{3 + 5/x^2}{2 - 1/x} = \frac{3 + 0}{2 - 0} = \frac{3}{2}$$

Both give $\dfrac{3}{2}$. Here dividing is just as quick.

6. (Core) Find $\displaystyle\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}$.

Solution

Both go to $\infty$, so the form is $\dfrac{\infty}{\infty}$.

$$\lim_{x \to \infty} \frac{1/x}{\dfrac{1}{2\sqrt{x}}} = \lim_{x \to \infty} \frac{2\sqrt{x}}{x} = \lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0$$

So $\sqrt{x}$ grows faster than $\ln x$.

7. (Core) The functions $f$ and $g$ are differentiable, with $f(3) = 0$, $g(3) = 0$, $f'(3) = 6$, and $g'(3) = -2$. Find $\displaystyle\lim_{x \to 3} \frac{f(x)}{g(x)}$, assuming $f'$ and $g'$ are continuous.

Solution

Since $f$ and $g$ are differentiable, they are continuous, so $\displaystyle\lim_{x \to 3} f(x) = f(3) = 0$ and $\displaystyle\lim_{x \to 3} g(x) = g(3) = 0$. The form is $\dfrac{0}{0}$, so L’Hospital’s Rule applies:

$$\lim_{x \to 3} \frac{f(x)}{g(x)} = \lim_{x \to 3} \frac{f'(x)}{g'(x)} = \frac{6}{-2} = -3$$

8. (Challenge) Find $\displaystyle\lim_{x \to 0} \frac{x - \sin x}{x^3}$.

Solution

Each step below is $\dfrac{0}{0}$ until the last:

$$\lim_{x \to 0} \frac{x - \sin x}{x^3} = \lim_{x \to 0} \frac{1 - \cos x}{3x^2} = \lim_{x \to 0} \frac{\sin x}{6x} = \lim_{x \to 0} \frac{\cos x}{6} = \frac{1}{6}$$

9. (Challenge) Find the value of the constant $k$ that makes $\displaystyle\lim_{x \to 0} \frac{e^{kx} - 1 - 2x}{x^2}$ a finite number, and find the value of the limit.

Solution

The form is $\dfrac{0}{0}$ for every $k$. Apply the rule once:

$$\lim_{x \to 0} \frac{ke^{kx} - 2}{2x}$$

The bottom goes to $0$, and the top goes to $k - 2$. If $k \ne 2$, this new limit doesn’t exist, but L’Hospital’s Rule only tells you something when the new limit does exist, so reason another way. By the rule, $\displaystyle\lim_{x \to 0} \frac{e^{kx} - 1 - 2x}{x} = \lim_{x \to 0} \frac{ke^{kx} - 2}{1} = k - 2$. So near $0$,

$$\frac{e^{kx} - 1 - 2x}{x^2} = \frac{e^{kx} - 1 - 2x}{x} \cdot \frac{1}{x} \approx \frac{k - 2}{x}$$

which blows up if $k \ne 2$. So the limit can only be finite if $k = 2$.

With $k = 2$ the form is $\dfrac{0}{0}$ again, so apply the rule a second time:

$$\lim_{x \to 0} \frac{4e^{2x}}{2} = 2$$

So $k = 2$, and the limit is $2$.