L'Hôpital's Rule/Proof 1

Proof
Let $l = \ds \lim_{x \mathop \to a^+} \frac {\map {f'} x}{\map {g'} x}$.

Let $\epsilon > 0$.

By the definition of limit, we ought to find a $\delta > 0$ such that:


 * $\forall x \in \R: \size {x - a} < \delta \implies \size {\dfrac {\map f x} {\map g x} - l} < \epsilon$

Fix $\delta$ such that:


 * $\forall x \in \R: \size {x - a} < \delta \implies \size {\dfrac {\map {f'} x} {\map {g'} x} - l} < \epsilon$

which is possible by the definition of limit.

Let $x$ be such that $\size {x - a} < \delta$.

By the Cauchy Mean Value Theorem with $b = x$:
 * $\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x - \map f a} {\map g x - \map g a}$

Since $\map f a = \map g a = 0$:
 * $\exists \xi \in \openint a z: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x} {\map g x}$

Now, as $a < \xi < x$, it follows that $\size{\xi - a} < \delta$ as well.

Therefore:


 * $\size {\dfrac {\map f x} {\map g x} - l} = \size {\dfrac {\map {f'} \xi} {\map {g'} \xi} - l} < \epsilon$

which leads us to the desired conclusion that:


 * $\ds \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$