L'Hôpital's Rule/Proof 2

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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

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.

$\blacksquare$


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