## L’Hopital’s Rule for Limits

1. Now we are concerned about limits and we can have nightmarish limits. For example if
(1a)$\displaystyle\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=0$
then
(1b)$\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{0}{0}$?
What can we do?

We end up using derivatives to help evaluate such limits. We have L’Hopital’s rule state, if $g'(c)\not=0$, then
(1c)$\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$

2. Sketch of Proof. We can use linear approximations writing
(2)$\displaystyle f(x+h)=f(x)+f'(x)h$ and $\displaystyle g(x+h)=g(x)+g'(x)h$
Use these approximations in (1a), then take the limit $h\to 0$. So our limit becomes
(3)$\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{h\to0}\lim_{x\to c}\frac{f(x)+f'(x)h}{g(x)+g'(x)h}$
We evaluate the inner limit first
(4)$\displaystyle\lim_{h\to0}\lim_{x\to c}\frac{f(x)+f'(x)h}{g(x)+g'(x)h}=\lim_{h\to0}\frac{f(c)+f'(c)h}{g(c)+g'(c)h}$
But look $f(c)=g(c)=0$, so (4) simplifies to
(5)$\displaystyle \lim_{h\to0}\frac{f(c)+f'(c)h}{g(c)+g'(c)h}= \lim_{h\to0}\frac{0+f'(c)h}{0+g'(c)h}$
Thus we obtain
(6)$\displaystyle \lim_{h\to0}\frac{f'(c)h}{g'(c)h}= \lim_{h\to0}\frac{f'(c)}{g'(c)}=\frac{f'(c)}{g'(c)}$
So we obtain
(7)$\displaystyle \lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$

3. Sketch of Proof for $\infty/\infty$. When
(8a)$\displaystyle\lim_{x\to c}f(x)=\pm\infty$
and
(8b)$\displaystyle\lim_{x\to c}g(x)=\pm\infty$
we run into similar problems. L’Hopital’s rule still holds. We just do the following trick:
(9)$\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{1/g(x)}{1/f(x)}$
and we recover the previous case…kind of!

Taking the derivative of top and bottom, we get
(10)$\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{-g'(x)/[g(x)^{2}]}{-f'(x)/[f(x)^{2}]}$
We can cross multiply, giving us
(11)$\displaystyle \frac{-g'(x)/[g(x)^{2}]}{-f'(x)/[f(x)^{2}]}=\frac{g'(x)[f(x)^{2}]}{f'(x)[g(x)^{2}]}$
Cross multiplying gives us
(12)$\displaystyle\lim_{x\to c}\frac{g(x)}{f(x)}=\lim_{x\to c}\frac{g'(x)}{f'(x)}$
and thus
(13)$\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$.

Notation: We will write
(14)$\displaystyle\lim_{x\to c}\frac{f(x)}{g(x)}\stackrel{L'H}{=}\lim_{x\to c}\frac{f'(x)}{g'(x)}$
when we apply L’Hopital’s rule.

Example 1. Consider the sequence
(15)$\displaystyle a_{n}=\frac{n^{2}}{\exp(n)}$.
What happens as $n\to\infty$? We have to apply L’Hopital’s rule twice
(16)\displaystyle \begin{aligned}a_{n}&=\frac{n^{2}}{\exp(n)}\\ &\stackrel{L'H}{=}\frac{2n}{\exp(n)}\\ &\stackrel{L'H}{=}\frac{2}{\exp(n)}\end{aligned}.
Thus we have
(17)$\displaystyle \lim_{n\to\infty}\frac{n^{2}}{\exp(n)}\stackrel{L'H}{=}\lim_{n\to\infty}\frac{2}{\exp(n)}=0$.