L'Hôpital's Rule/Proof 2

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Let $f$ and $g$ be real functions which are continuous on the closed interval $\left[{a \,.\,.\, b}\right]$ and differentiable on the open interval $\left({a \,.\,.\, b}\right)$.


$\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$


$f \left({a}\right) = g \left({a}\right) = 0$


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

provided that the second limit exists.


Take the Cauchy Mean Value Theorem with $b = x$:

$\displaystyle \exists \xi \in \left({a \,.\,.\, x}\right): \frac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \frac {f \left({x}\right) - f \left({a}\right)} {g \left({x}\right) - g \left({a}\right)}$

Then if $f \left({a}\right) = g \left({a}\right) = 0$ we have:

$\displaystyle \exists \xi \in \left({a \,.\,.\, x}\right): \frac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \frac {f \left({x}\right)} {g \left({x}\right)}$

Note that $\xi$ depends on $x$; that is, different values of $x$ may require different values of $\xi$ to make the above statement valid.

It follows from Limit of Function in Interval that $\xi \to a$ as $x \to a$.

Also, $\xi \ne a$ when $x > a$.

So from Hypothesis 2 of Limit of Composite Function, it follows that:

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

Hence the result.