L'Hôpital's Rule/Proof 1

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

Let $\epsilon \in \R_{>0}$.

By the definition of limit, we ought to find a $\delta \in \R_{>0}$ 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.

Define:
 * $\map {f_0} x = \begin{cases}

\map f x & : x \in \openint a b \\ 0 & : x = a \end{cases}$
 * $\map {g_0} x = \begin{cases}

\map f x & : x \in \openint a b \\ 0 & : x = a \end{cases}$

By definition of right-continuous, it follows that $f_0$ and $g_0$ are continuous on $\hointr a b$.

Let $x_\delta$ be such that $0 < x_\delta - a < \delta$.

We have that $f_0$ and $g_0$ are continuous on $\closedint a {x_\delta}$, and differentiable on $\openint a {x_\delta}$

Thus, by the Cauchy Mean Value Theorem with $b = x_\delta$:
 * $\exists \xi \in \openint a {x_\delta}: \dfrac {\map {f'_0} \xi} {\map {g'_0} \xi} = \dfrac {\map {f_0} {x_\delta} - \map {f_0} a} {\map {g_0} {x_\delta} - \map {g_0} a}$

Since $\map {f_0} a = \map {g_0} a = 0$:
 * $\exists \xi \in \openint a {x_\delta}: \dfrac {\map {f'_0} \xi} {\map {g'_0} \xi} = \dfrac {\map {f_0} {x_\delta}} {\map {g_0} {x_\delta}}$

But since $\xi, x_\delta \in \openint a b$:
 * $\map {f'_0} \xi = \map {f'} \xi$
 * $\map {g'_0} \xi = \map {g'} \xi$
 * $\map {f_0} {x_\delta} = \map f {x_\delta}$
 * $\map {g_0} {x_\delta} = \map g {x_\delta}$

Therefore:
 * $\exists \xi \in \openint a {x_\delta}: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f {x_\delta}} {\map g {x_\delta}}$

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

Therefore:


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

which leads us to the desired conclusion that:


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