Let $f$ and $g$ be differentiable on the interval $a\le x <b $, with $g'(x) \ne 0$ there. If
$\lim_{x\to b} f(x)= \infty$ and $\lim_{x\to b} g(x) = \infty $
and if $\lim_{x\to b} \frac {f'(x)}{g'(x)} = L $
then $\lim \frac {f(x)}{g(x)} = L $
proof. We start from the fact that $f(x) \rightarrow \infty $, $g(x) \rightarrow \infty $, and $\frac {f'(x)}{g'(x)} \rightarrow L $ as $x $ approaches $b $. Since $g'(x) $ is never $0$, we know that it must be positive and thus that $g $ is strictly increasing; we may therefore assume $g (x)>0$ for all $x \in [a,b]$. Given $\epsilon >0$, choose $x_0$ so that if $x_0 <t <b$,
$-\epsilon < \frac {f'(t)}{g'(t)} -L < \epsilon$
Since $g'(t)>0$, we may rewrite this as
(3-5) $(L-\epsilon)g'(t) < f'(t) < (L + \epsilon)g'(t) $
The right half can be written
$f'(t) - (L+ \epsilon)g'(t) < 0$
which implies that the function $f (x) - (L + \epsilon)g (x)$ is decreasing, since its derivative is negative on $[x_0,b] $. A decreasing function is necessarily bounded above, so that we have $f (x) - (L - \epsilon)g (x) < B $. * possible typo
Divide by $g (x) $, which is positive, and arrive at
$\frac {f (x)}{g (x)} < L + \epsilon + \frac{B}{g (x)} $
Since $g (x) \rightarrow \infty $ as $x \rightarrow b $, we see that for some $x_1$ near $b $, $\frac {f (x)}{g (x)} < L + 2\epsilon$ for all $x, x_1 <x <b $.
If we return to (3-5) and carry out a similar calculation using the left half of the inequality, we obtain $\frac {f (x)}{g(x)} > L - 2\epsilon $, and the combination proves that $\lim_{x\to b } \frac {f (x)}{g (x)} $
I wanted to know where the $2\epsilon $ comes from? Do we just assume that $\epsilon = \frac {B}{g (x)} $?
Also, what happens if $\lim_{x \to b} f (x) =0$ and $\lim_{x\to b}g (x)=\infty $? Can we still apply L'Hospital's Rule?
Sunce $\lim_{x\to b}g(x)=\infty$, $\lim_{x\to b}\frac B{g(x)}=0$ and therefore, if $x_1$ is close enough to $b$, then $\frac B{g(x)}<\varepsilon$.
And if $\lim_{x\to b}f(x)=0$ and $\lim_{x\to b}g(x)=\infty$, then $\lim_{x\to b}\frac{f(x)}{g(x)}=0$.