L'Hôpital's Rule/Proof 1

Proof
Let $l = \displaystyle \lim_{x \mathop \to a^+} \frac{f' \left({x}\right)}{g' \left({x}\right)}$.

Let $\epsilon > 0$.

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


 * $\forall x: \left\vert{x - a}\right\vert < \delta \implies \left\vert{\dfrac {f \left({x}\right)} {g \left({x}\right)} - l}\right\vert < \epsilon$

Fix $\delta$ such that:


 * $\forall x: \left\vert{x - a}\right\vert < \delta \implies \left\vert{\dfrac {f' \left({x}\right)} {g' \left({x}\right)} - l}\right\vert < \epsilon$

which is possible by the definition of limit.

Let $x$ be such that $\left\vert{x - a}\right\vert < \delta$.

By the Cauchy Mean Value Theorem with $b = x$:
 * $\exists \xi \in \left({a \,.\,.\, x}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({x}\right) - f \left({a}\right)} {g \left({x}\right) - g \left({a}\right)}$

Since $f \left({a}\right) = g \left({a}\right) = 0$, we have:
 * $\exists \xi \in \left({a \,.\,.\, x}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({x}\right)} {g \left({x}\right)}$

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

Therefore:


 * $\left\vert{\dfrac {f \left({x}\right)} {g \left({x}\right)} - l }\right\vert = \left\vert{ \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} - l}\right\vert < \epsilon$

which leads us to the desired conclusion that:


 * $\displaystyle \lim_{x \mathop \to a^+} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \mathop \to a^+} \frac {f' \left({x}\right)} {g' \left({x}\right)}$