4.4: L’Hopital’s Rule

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

L’Hopital’s Rule Let $f(x)$ and $g(x)$ be differentiable functions with $\lim _{x \to a} f(x) = 0 \text { and } \lim _{x \to a} g(x) = 0,$ or $\lim _{x \to a} f(x) = \infty \text { and } \lim _{x \to a} g(x) = \infty .$ If $g'(x) \neq 0$ near $a$ , 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 $\infty$ or $-\infty$ ).

Evaluate $\displaystyle \lim _{x \to 0} \frac {1-\cos (x)}{x}$ .

We first try plugging in $x=0$ into the function. Let $f(x) = 1-\cos (x)$ and $g(x) = x$ . Then $f(0) = \answer {0}$ and $g(x) = \answer {0}$ , so we get a $0/0$ limit. We will use L’Hopital’s rule to evaluate the limit. Notice $f'(x) = \answer {\sin (x)},\quad g'(x) = \answer {1},$ so L’Hopital’s Rule says $\lim _{x \to 0} \frac {f(x)}{g(x)} = \lim _{x \to 0} \frac {\sin (x)}{1}.$ The limit on the right hand side is equal to $\answer {0}$ , so our original limit is also $\answer {0}$ .

Evaluate $\displaystyle \lim _{x \to \infty } \frac {e^x+1}{e^{2x}-5}$ .

We first try plugging in $x=0$ into the function. Let $f(x) = e^x+1$ and $g(x) = e^{2x}-5$ . Then $\lim _{x \to \infty } f(x) = \answer {\infty }$ and $\lim _{x \to \infty } g(x) = \answer {\infty }$ , so we get a $\infty /\infty$ limit. We will use L’Hopital’s rule to evaluate the limit. Notice $f'(x) = \answer {e^x},\quad g'(x) = \answer {2e^{2x}},$ so L’Hopital’s Rule says $\lim _{x \to \infty } \frac {f(x)}{g(x)} = \lim _{x \to \infty } \frac {e^x}{2e^{2x}}.$ Technically, this is still $\infty /\infty$ , so we could use L’Hopital’s rule. But you’ll quickly find that it doesn’t lead anywhere. Instead, notice that $\frac {e^x}{2e^{2x}} = \frac {1}{\answer {2e^x}},$ and $\lim _{x \to \infty } \frac {1}{2e^x} = \answer {0},$ so by L’Hopital’s rule and this algebra, we get a limit of $\answer {0}$ .

For the remaining problems, evaluate the limit. Note that not all of them will require L’Hopital’s rule.

The limit $\lim _{x \to 3} \frac {\sin (\pi x)}{x-3} = \answer {-\pi }$ .

The limit $\lim _{x \to \infty } \frac {e^x}{x^2} = \answer {\infty }$ .

The limit $\lim _{x \to 0} \frac {e^x-1}{x} = \answer {1}$ .

The limit $\lim _{x \to 3} \frac {\sin (\pi x)}{x} = \answer {0}$ .

The limit $\lim _{x \to \pi /4} \frac {\tan (x)-1}{\tan (3x)+1} = \answer {1/3}$ .