Please can you check my proof of de l'Hospital's rule Statement: If $f,g : (a,b) \to \mathbb R$ are differentiable and $\lim_{x \to a^+}f(x) = \lim_{x \to a^+} g(x) = 0 $ and $\lim_{x \to a^+} {f'(x) \over g'(x)} = L$ then $\lim_{x \to a^+} {f(x) \over g(x)} = L$.
Proof: Let $\varepsilon > 0$. The goal is to find $\delta >0$ such that $0 < |x-a|< \delta$ implies $|{f(x) \over g(x)}-L| < \varepsilon$. Since $\lim_{x \to a^+} {f'(x) \over g'(x)} = L$  there is $\delta'>0$ such that $0 < |x-a|<\delta'$ implies $|{f'(x) \over g'(x)}-L| < \varepsilon$. By the mean value theorem there exists $c \in (a,a+\delta')$ such that 
$$ {f'(c) \over g'(c) } = {f(a + \delta') -f(a) \over g(a+\delta') -g(a) } = {f(a + \delta') \over g(a+\delta')}$$
That is, for every $\delta_n$ there exists $c_n \in (a,a+\delta_n)$ such that ${f'(c) \over g'(c)} ={f(a+\delta_n) \over g(a + \delta_n)}$. Hence if $\delta< \delta'$ then for every $x \in (a,a+\delta)$:
$$ \left | {f(x) \over g(x)} -L \right | =  \left | {f'(c) \over g'(c)} -L \right | < \varepsilon$$
 A: While the line of reasoning is fine in general
i) I'd suggest you mention that you use the not so well known Cauchy variant of the MVT, 
ii) you don't get equality for the two  differences in the last formuala with display style for the given $x$ and 
iii) the formulation 'that is, for every $\delta_n$... is confusing at first reading, at least for me. If you write: 'that is, if $0<\delta_n<\delta^\prime$...' I'd consider this to be easier to read. This point is nitpicking, however.
A: Proof: Let $\varepsilon > 0$. The goal is to find $\delta >0$ such that $0 < |x-a|< \delta$ implies $\left |{f(x) \over g(x)}-L \right | < \varepsilon$. Define $f(a) = g(a) = 0$. Then $f,g$ are continuous on $[a,b)$. Since $\lim_{x \to a^+} {f'(x) \over g'(x)} = L$  there is $\delta'>0$ such that $0 < |x-a|< \delta'$ implies $\left | {f'(x) \over g'(x)}-L\right | < \varepsilon$. For every such $x$ there exists $\delta > 0$ such that $x = a + \delta$ and by the mean value theorem there exists $c \in (a,a+\delta) = (a,x)$ such that 
$$ {f'(c) \over g'(c) } = {f(a + \delta) -f(a) \over g(a+\delta) -g(a) } = {f(a + \delta) \over g(a+\delta)} = {f(x) \over g(x)}$$
Then for every $x$ with  $0< |x-a|<\delta'$,
$$ \left | {f(x) \over g(x)} -L \right | =  \left | {f(a+\delta) \over g(a + \delta)} -L \right | =  \left |  {f'(c) \over g'(c) }  -L \right | < \varepsilon$$
