Given: lim x$\rightarrow x_0$ f(x) = $\infty$ and lim x$\rightarrow x_0$ g(x) = L>0.
Prove that lim x$\rightarrow x_0$ $(g(x)/f(x))$ = 0.
My attempt:
Suppose that lim f(x)=$\infty$ and lim g(x) = L > 0.
Since lim f(x) = $\infty$, we know that if $\forall$A, $\exists \delta_1$>0, if 0<$|x-x_0|$<$\delta_1$, then $f(x)>A.$
So, $1/f(x)$$\leq$$1/A$ holds as long as $|x-x_0|<\delta$.
Further, lim$g(x)$=L>0, if $\forall$A, $\exists \delta_2$>0, if 0<$|x-x_0|$<$\delta_2$, then $|g(x)-L|$<$\epsilon$
So, we have $L-\epsilon<g(x)<L+\epsilon$.
Let $\epsilon = \epsilon_2A-L$, then we get that $g(x)<L+\epsilon_2A-L=\epsilon A$.
Using Limit Properties, we finally have: $lim g(x)/lim f(x) < (\epsilon_2A)/A$=$\epsilon$
QED
I was hoping to get some input on my proof. I know it needs more work, but I'm not sure where and how to improve it at this point.
Just to make things easier, since $g(x)\to L$ when $x\to x_0$, there is $\delta >0$ and $M>0$ s.t. $|g(x)|\leq M$ for all $|x-x_0|<\delta $. Therefore, $$\left|\frac{g(x)}{f(x)}\right|\leq \frac{|M|}{|f(x)|},\quad \text{if }|x-x_0|<\delta.$$
If $\varepsilon >0$, take $\tilde \delta \in (0,\delta )$ s.t. $|f(x)|>\frac{M}{\varepsilon }$, and you get $$\left|\frac{g(x)}{f(x)}\right|<\varepsilon ,$$ whenever $|x-x_0|<\tilde \delta $. The claim follows.