• Recognize limits that produce indeterminate forms.
  • Apply L’Hôpital’s Rule to evaluate a limit.

Assignment

  • Vocabulary and teal boxes
  • p513 1, 4, 5, 8–11, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 65 69–72, 80–82, 101, 102, 113–115

Additional Resources

Indeterminate Forms

When trying to find a limit through direct substitution (yes, we’re going all the way back to limits), you sometimes get strange results.

\[\begin{align} \lim_{x\to 0} \frac{x}{x} \qquad \lim_{x\to 0} \frac{x^2}{x} \qquad \lim_{x\to 0} \frac{x}{x^3} \end{align}\]

Each limit, if using direct substitution, evaluates to $\frac{0}{0}$, but their actual limits are

\[\begin{align} \lim_{x\to 0} \frac{x}{x} &= 1 \\ \lim_{x\to 0} \frac{x^2}{x} &= 0 \\ \lim_{x\to 0} \frac{x}{x^3} &= \infty \end{align}\]

$\frac{0}{0}$ is an example of an indeterminate form. Conflicting rules make it difficult to determine the actual limit. Does it reduce to 1? Is it 0? Does it go to infinity? Keep in mind that limits are about the approach to a number, not the number itself. With the limits above, you have two competing functions, your numerator and denominator, both trying to approach 0. Who gets there first, and what the other one is doing in the meantime, will determine the actual limit.

The two indeterminate forms you need to be aware of are $\frac{0}{0}$ and $\frac{\infty}{\infty}$. There are others, but they do no appear on the AP exam.

L’Hôpital’s Rule

Luckily, dealing with these two indeterminate forms is easy. When finding a limit, if direct substitution yields $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then you can use L’Hôpital’s rule.

L’Hôpital’s Rule (abbreviated)

If the limit of $f(x)/g(x)$ as $x$ approaches $c$ produces the indeterminate form $0/0$ or $\infty/\infty$, then

\[\begin{align} \lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)} \end{align}\]

Find the derivative of both the numerator and denominator, and try direct substitution again. Was the result another indeterminate form? L’Hôpital’s rule again.

Example 1

Evaluate $\displaystyle \lim_{x\to0}\frac{e^{2x}-1}{x}$.

Direct substitution yields $\frac{0}{0}$, so we can use the rule.

\[\begin{align} \lim_{x\to0}\frac{e^{2x}-1}{x} &= \lim_{x\to0}\frac{\frac{d}{dx}[e^{2x}-1]}{\frac{d}{dx}[x]}\\[1em] &=\lim_{x\to0}\frac{2e^{2x}}{1} \\ &= 2 \end{align}\]

$\blacksquare$

Example 2

Evaluate $\displaystyle \lim_{x\to\infty}\frac{\ln x}{x}$.

This time we get $\frac{\infty}{\infty}$. Rule time again.

\[\begin{align} \lim_{x\to\infty}\frac{\ln x}{x} &= \lim_{x\to\infty}\frac{\frac{1}{x}}{1} \\ &= \lim_{x\to\infty} \frac{1}{x} \\ &= 0 \end{align}\]

$\blacksquare$

Example 3

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

We get $\frac{\infty}{\infty}$, so rule time.

\[\begin{align} \lim_{x\to-\infty}\frac{x^2}{e^{-x}} &= \lim_{x\to-\infty}\frac{2x}{-e^{-x}} \end{align}\]

Oh, now it’s $\frac{-\infty}{-\infty}$. Negatives don’t matter. It’s still indeterminate and therefore a candidate for the rule. Run it through again.

\[\begin{align} \lim_{x\to-\infty}\frac{2x}{-e^{-x}} &= \lim_{x\to-\infty}\frac{2}{e^{-x}} \\ &= 0 \end{align}\]

$\blacksquare$