Hospital/Differentiable in inner interval/Fact/Proof

Proof

Because ${}g'$ has no zero in the interval and ${}g(a)=0$ holds, it follows, because of fact, that ${}a$ is the only zero of ${}g$. Let ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ denote a sequence in ${}I\setminus \{a\}$, converging to ${}a$.

For every ${}x_{n}$ there exists, due to fact, applied to the interval ${}I_{n}:=[x_{n},a]$ or ${}[a,x_{n}]$, a ${}c_{n}$ (in the interior of ${}I_{n}$,) fulfilling

${}{\frac {f(x_{n})-f(a)}{g(x_{n})-g(a)}}={\frac {f'(c_{n})}{g'(c_{n})}}\,.$

The sequence ${}{\left(c_{n}\right)}_{n\in \mathbb {N} }$ converges also to ${}a$, so that, because of the condition, the right-hand side converges to ${}{\frac {f'(a)}{g'(a)}}=w$. Therefore, also the left-hand side converges to ${}w$, and, because of ${}f(a)=g(a)=0$, this means that ${}{\frac {f(x_{n})}{g(x_{n})}}$ converges to ${}w$.

To fact