L'Hopital's Rule

L'Hô pital's Rule can be strengthened to include the case when $g^{\prime}(a)=0$ and the indeterminate form " $\infty/\infty$ ", the case when both $\mathrm{f}$ and $\mathrm{g}$ increase without any bound.

Example 2: Evaluate $\lim \limits_{x \rightarrow \infty} \frac{e^{7 x}}{5 x}$.

Solution: As " $x \rightarrow \infty$ ", both $f(x)=e^{7 x}$ and $g(x)=5 x$ increase without bound so we have an " $\infty / \infty$ " indeterminate form and can use the Strong Version l'Hô pital's Rule:

$\lim \limits_{x \rightarrow \infty} \frac{e^{7 x}}{5 x}=\lim \limits_{x \rightarrow \infty} \frac{7 e^{7 x}}{5}=\infty$

The limit of $f '/g '$ may also be an indeterminate form, and then we can apply l'Hô pital's Rule to the ratio $f ' / \mathrm{g} '$. We can continue using l'Hô pital's Rule at each stage as long as we have an indeterminate quotient.

Example 3: $\lim \limits_{x \rightarrow 0} \frac{x^{3}}{x-\sin (x)}$

Solution: As $x \rightarrow 0, f(x)=x^{3} \rightarrow 0$ and $g(x)=x-\sin (x) \rightarrow 0$ so

$\lim \limits_{x \rightarrow 0} \frac{x^{3}}{x-\sin (x)}=\lim \limits_{x \rightarrow 0} \frac{3 x^{2}}{1-\cos (x)} \rightarrow$ " $\frac{0}{0}$ " so we can use l'Hô pital's Rule again

=$\lim \limits_{x \rightarrow 0} \frac{6 x}{\sin (x)} \rightarrow$ " $\frac{0}{0}$ " and again

=$\lim \limits_{x \rightarrow 0} \frac{6}{\cos (x)}=\frac{6}{1}=6$.

Practice 2: Use l'Hô pital's Rule to find $\lim \limits_{x \rightarrow \infty} \frac{x^{2}+e^{x}}{x^{3}+8 x}$.