# Hospital/Differentiable in inner interval/Fact/Proof

Because ${\displaystyle {}g'}$ has no zero in the interval and ${\displaystyle {}g(a)=0}$ holds, it follows, because of fact, that ${\displaystyle {}a}$ is the only zero of ${\displaystyle {}g}$. Let ${\displaystyle {}{\left(x_{n}\right)}_{n\in \mathbb {N} }}$ denote a sequence in ${\displaystyle {}I\setminus \{a\}}$, converging to ${\displaystyle {}a}$.
For every ${\displaystyle {}x_{n}}$ there exists, due to fact, applied to the interval ${\displaystyle {}I_{n}:=[x_{n},a]}$ or ${\displaystyle {}[a,x_{n}]}$, a ${\displaystyle {}c_{n}}$ (in the interior[1] of ${\displaystyle {}I_{n}}$,) fulfilling
${\displaystyle {}{\frac {f(x_{n})-f(a)}{g(x_{n})-g(a)}}={\frac {f'(c_{n})}{g'(c_{n})}}\,.}$
The sequence ${\displaystyle {}{\left(c_{n}\right)}_{n\in \mathbb {N} }}$ converges also to ${\displaystyle {}a}$, so that, because of the condition, the right-hand side converges to ${\displaystyle {}{\frac {f'(a)}{g'(a)}}=w}$. Therefore, also the left-hand side converges to ${\displaystyle {}w}$, and, because of ${\displaystyle {}f(a)=g(a)=0}$, this means that ${\displaystyle {}{\frac {f(x_{n})}{g(x_{n})}}}$ converges to ${\displaystyle {}w}$.
1. The interior of a real interval ${\displaystyle {}I\subseteq \mathbb {R} }$