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8-2 L’Hopital’s Rule

Learning Targets

You will be able to

Concepts / Definitions

How do we algebraically calculate $\lim_{x\to 1} \frac{\ln x}{x-1}$?

In the past, we’ve looked at the graph.

Recall previous methods for calculating limits.

Comparing two points

L’Hopital’s Rule

L’Hopital’s rule is used for calculating limits of the form $\frac 00$ or $\frac \infty\infty$.

Suppose that $f(a) = g(a) = 0$, and that $f$ and $g$ are differentiable on an open interval $I$ containing $a$, and that $g’(x) \neq 0$ on $I$ if $x \neq a$. Then, if the limit below exists,

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

Deriving L’Hopital’s Rule

If you look very closely at the graph (zoom in!), the equation on either side of the graph looks linear.


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

$\lim_{x\to a} = \frac{f(x)-0}{g(x)-0}$

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

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

$\lim_{x\to a} = \frac{f’(a)}{g’(a)}$


Determine $\lim_{x\to\infty} (x\sin\frac 1x)$

Determine $\lim_{x\to 0} (\cot x + \frac 1x)$

$\lim_{x\to 0} (\frac{\cos x}{\sin x} + \frac 1x)$

$\lim_{x\to 0} (\frac{\cos x}{\sin x} \frac{\sin x}{x} + \frac 1x)$

$\lim_{x\to 0} (\frac{\cos x}{x} + \frac 1x)$

$\lim_{x\to 0} (\frac{\cos x + 1}{x})$