L'Hôpital's Rule/Proof 1
Theorem
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Let:
- $\forall x \in \openint a b: \map {g'} x \ne 0$
where $g'$ denotes the derivative of $g$ with respect to $x$.
Let:
- $\map f a = \map g a = 0$
Then:
- $\displaystyle \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$
provided that the second limit exists.
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:
- $\displaystyle \lim_{x \mathop \to a^+} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \mathop \to a^+} \frac {f' \left({x}\right)} {g' \left({x}\right)}$
$\blacksquare$