L'Hospital's Rule

In Unit 1 you met limits like $\lim_{x \to 0}\frac{\sin x}{x}$, where both the top and the bottom approach 0. The form $\frac{0}{0}$ tells you nothing by itself: the limit might be 1, or 7, or not exist. Back then you needed algebra tricks or a squeeze. Now that you have derivatives, one rule handles most of these limits, and it comes from the idea you just used for tangent line approximations.

Why slopes decide a $\frac{0}{0}$ limit

Suppose $f(a) = 0$ and $g(a) = 0$, both differentiable. Near $a$, each function is close to its tangent line:

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

So their quotient is close to $\frac{f'(a)(x - a)}{g'(a)(x - a)}$$= \frac{f'(a)}{g'(a)}$. Both functions shrink to 0, but at rates set by their slopes, and the ratio of the slopes is what survives.

L'Hospital's rule

If $\lim_{x \to a}f(x) = 0$ and $\lim_{x \to a}g(x) = 0$, or if both limits are infinite, 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 infinite). The same holds for limits as $x \to \infty$. Three warnings:

• Check the form first. The rule needs $\frac{0}{0}$ or $\frac{\infty}{\infty}$. On the exam, show it: write both limits.
• Differentiate the top and the bottom separately. This is not the quotient rule.
• Never write "$= \frac{0}{0}$". It is a form, not a number. Write "$\lim(1 - \cos x) = 0$ and $\lim x^2 = 0$" instead.

When the form is not indeterminate

Using the rule on a limit that is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ gives wrong answers. For $\lim_{x \to 1}\frac{x^2 + 1}{x + 1}$, the form is $\frac{2}{2}$, so the limit is simply 1. Blindly differentiating gives $\lim\frac{2x}{1} = 2$, which is wrong. A check of the form takes five seconds and saves the point.

From a table of values

The form $\frac{\infty}{\infty}$: comparing growth

As $x \to \infty$, the rule compares how fast two functions grow.

Slips that cost points

• Applying the rule without showing that both limits are 0 (or both infinite).
• Using the quotient rule instead of differentiating top and bottom separately.
• Writing $\frac{0}{0} =$ a number, or "$= \frac{0}{0}$" as if it were a value.
• Forgetting the chain rule inside: $\frac{d}{dx}\sin(5x)$$= 5\cos(5x)$.
• Stopping after one round when the new limit is still indeterminate.