L'Hôpital's Rule

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

Suppose that $$\exists x \in \left({a \, . \, . \, b}\right): g^{\prime} \left({x}\right) \ne 0$$.

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

Then $$\lim_{x \to a^+} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \to a^+} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$$

provided that the second limit exists.

Corollary 1
Suppose that instead of $$f \left({a}\right) = g \left({a}\right) = 0$$, we have that $$\exists c \in \left({a \, . \, . \, b}\right): f \left({c}\right) = g \left({c}\right) = 0$$.

Then $$\lim_{x \to c} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \to c} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$$

Corollary 2
Suppose that instead of $$f \left({a}\right) = g \left({a}\right) = 0$$, we have that $$f \left({x}\right) \to \infty$$ and $$g \left({x}\right) \to \infty$$ as $$x \to a^+$$.

Then $$\lim_{x \to a^+} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \to a^+} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$$

provided that the second limit exists.

Proof
Take the Cauchy Mean Value Theorem with $$b = x$$: $$\exists \xi \in \left({a \, . \, . \, x}\right): \frac {f^{\prime} \left({\xi}\right)} {g^{\prime} \left({\xi}\right)} = \frac {f \left({x}\right) - f \left({a}\right)} {g \left({x}\right) - g \left({a}\right)}$$.

Then if $$f \left({a}\right) = g \left({a}\right) = 0$$ we have $$\exists \xi \in \left({a \, . \, . \, x}\right): \frac {f^{\prime} \left({\xi}\right)} {g^{\prime} \left({\xi}\right)} = \frac {f \left({x}\right)} {g \left({x}\right)}$$.

Note that $$\xi$$ depends on $$x$$, i.e. $$\xi$$ is a function of $$x$$.

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 $$\lim_{x \to a^+} \frac {f^{\prime} \left({\xi}\right)} {g^{\prime} \left({\xi}\right)} = \lim_{x \to a^+} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)} $$.

Hence the result.

Proof of Corollary 1
This follows directly from the definition of limit.

If $$\lim_{x \to c} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$$ exists, it follows that $$\lim_{x \to c} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)} = \lim_{x \to c^+} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$$.

That is, if there exists such a limit, it is also a limit from the right.

Proof of Corollary 2
We have that $$f \left({x}\right) \to \infty$$ and $$g \left({x}\right) \to \infty$$ as $$x \to a^+$$.

Thus it follows that $$\frac 1 {f \left({x}\right)} \to 0$$ and $$\frac 1 {g \left({x}\right)} \to 0$$ as $$x \to a^+$$.

The result follows, after some algebra.