L'Hôpital's Rule/Corollary 2

Corollary to L'Hôpital's Rule
Let $f$ and $g$ be real functions which are differentiable on the open interval $\openint a b$.

Suppose that $\forall x \in \openint a b: \map {g'} x \ne 0$. Suppose that $\map f x \to \infty$ and $\map g x \to \infty$ as $x \to a^+$.

Then:
 * $\ds \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$

provided that the second limit exists.

Proof
Let $\ds \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x} = L$.

Let $\sequence {x_n}$ be a sequence such that:
 * $\quad x_n \in \openint a b$ for all $n \in \N$ and $\ds \lim_{n \mathop \to \infty} x_n = a$

From Intermediate Value Theorem for Derivatives and the definition of limit of real function, it follows that:
 * $\ds \lim_{n \mathop \to \infty} \map g {x_n} = \infty$

and $\sequence {\map g {x_n} }$ is strictly increasing.

Consider the range $\closedint {x_{n - 1} } {x_n} \subset \openint a b$ where $n \ge 2$.

By Cauchy Mean Value Theorem, there exists $c_n \in \openint {x_{n - 1} } {x_n}$ such that:
 * $\ds \frac {\map f {x_n} - \map f {x_{n - 1} } } {\map g {x_n} - \map g {x_{n - 1} } } = \frac {\map {f'} {c_n} } {\map {g'} {c_n} }$

From the above and the Squeeze Theorem for Real Sequences:
 * $\ds \lim_{n \mathop \to \infty} c_n = a$

and:
 * $\ds \lim_{n \mathop \to \infty} \frac {\map f {x_n} - \map f {x_{n - 1} } } {\map g {x_n} - \map g {x_{n - 1} } } = \lim_{n \mathop \to \infty} \frac {\map {f'} {c_n} } {\map {g'} {c_n} } = L$

So, by Stolz-Cesàro Theorem:
 * $\ds \lim_{n \mathop \to \infty} \frac {\map f {x_n} } {\map g {x_n} } = L$

From the definition of limit of real function, we deduce that:
 * $\ds \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = L = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$

Remarks

 * The proof does not actually use the assumption $\ds \lim_{x \mathop \to a^+} \map f x = \infty$.


 * Cases $x \to b^-$, $x \to \pm \infty$ and $\map g x \to -\infty$ can be proven similarly.