How to prove L'Hospital's rule using $\varepsilon$-$\delta$ method? For $\varepsilon$-$\delta$ proofs, basically we need to find a $\delta$ such that $|F(x)-L|<\epsilon$ whenever, $0<|x-a|<\delta$ (for a small positive number $\epsilon$).
To prove L'Hospital's rule (for when numerator and denominator function both tend to $0$ as $x\rightarrow a^{+}$)  let us assume $F(x)=\frac{f(x)}{g(x)}$. Where, $\lim_{x\rightarrow a^{+}}f(x)=0$ and $\lim_{x\rightarrow a^{+}}g(x)=0$. 
I claim that $L=\lim_{x \rightarrow a^{+}}\frac{f'(x)}{g'(x)}$. Now I need to prove this $L$ is indeed the limit.
$$\left|\frac{f(x)}{g(x)}-\lim_{x \rightarrow a^{+}}\frac{f'(x)}{g'(x)}\right|<\epsilon.$$
But after this I cannot understand how to find $\delta$ as a function of $\epsilon$, so that I can complete the proof. How should I proceed?
 A: You need to assume that $$\lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}$$ exists. This means that for any $\varepsilon > 0 $ there is $\delta > 0 $ such that 
$$\tag{$\ast$}\quad\quad\quad\quad\left| \frac{f^\prime(y)}{g^\prime(y)}- \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}\right|< \varepsilon$$
whenever $|y-a|< \delta$.
Now the Cauchy version of the mean value theorem states that, if $f$ and $g$ are differentiable on $[b,a]$, say, and $x\in [b,a]$, and if $g(x)-g(a)\neq 0$, then there exists $c\in (x, a)$ such that 
$$\frac{f(x)-f(a)}{g(x)-g(a) }=\frac{f^\prime(c)}{g^\prime(c)}$$
I assume this to be known, the tricky part is to be able to choose $c$ for both the numerator and denominator simultaneously. This also is true only if $a\neq \infty$, so the following proof of l'Hopital only works under this additional assumption. 
Now let's look at the claim. Choose $\varepsilon > 0$. You want to estimate 
$$\left| \frac{f(y)}{g(y)}- \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}\right| < \varepsilon$$ for $y$ close to $a$. Choose $\delta >0 $ such that $(*) $ is true for your choice of $\varepsilon$.
 By the mean value theorem (Cauchy version) and the assumption $f(a)=g(a) =0$, the difference you are looking at is just the same as
$$\left| \frac{f^\prime(c)}{g^\prime(c)}- \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}\right|$$ 
for some $c\in (y,a)$ which is $<\varepsilon$ by $(*)$ if only $|y-a| < \delta$.
