Suppose that $f$ and $g$ are differentiable functions on an open interval $I$ and that $p \in I$. If $\lim_{x \to p} f(x) = \lim_{x \to p} g(x) = 0$ and if
$$\lim_{x \to p} \frac{f'(x)}{g'(x)}$$
exists and equals a real number $l$ then
$$\lim_{x \to p} \frac{f(x)}{g(x)}=l$$
Proof: Fix a real number $a > l$. By the definition of limit there is a number $q > p$ such that, if $p < x< q$, then $ \frac{f'(x)}{g'(x)}<a$. I don't know how to derive this.
You have $\lim_{x\to p}\big( \text{something}\big) = \ell.$
And you have $a>\ell.$
That implies that there is some open interval about $p$ for which, if $x$ is in that open interval and $x\ne p,$ then $\big(\text{something}\big) <a. $
Similarly if $b<\ell$ then there is some open interval about $p$ for which, if $x$ is in that open interval and $x\ne p,$ then $\big(\text{something})>b.$
This is a consequence of the definition of limit: No matter how small $\varepsilon>0$ is, there is some open interval about $p$ such that if $x$ is in that open interval and $x\ne p,$ then $\big(\text{something}\big)$ differs from $\ell$ by less than $\varepsilon.$ Generally as $\varepsilon$ gets smaller, that open interval about $p$ must get smaller.
