> [!Theorem] Theorem (l'Hopital's rule at a point) > Let $f,g:I\to \mathbb{R}$ be [[Fréchet Differentiation|differentiable real functions]] on a [[Open Real Interval|open interval]] $I.$ If there exists $c\in I$ such that $f(c)=g(c)=0$ then $\lim_{ x \to c } \frac{f(x)}{g(x)} = \lim_{ x \to c } \frac{f'(x)}{g'(x)}$provided the second [[Limit of Real Function at a Point|limit]] exists. **Proof**. Suppose that $\lim_{ x \to c } \frac{f'(x)}{g'(x)} $does indeed exist. Then it cannot be that $g'(x)=0$ at a sequence of points converging to $c$. So there is some interval around $c$ on which $g'$ is non-zero. This enable us to apply [[Cauchy Mean Value Theorem]]. Because $f(c)=g(c)=0$ $\lim_{ x \to c } \frac{f(x)}{g(x)} = \lim_{ x \to c } \frac{f(x)-f(c)}{g(x)-g(c)} $ As long as $x$ is in the region around $c$ where $g' \neq 0,$ by Cauchy MVT, there is a point $t$ between $c$ and $x$ where $\frac{f(x)-f(c)}{g(x)-g(x)} = \frac{f'(t)}{g'(t)}$As $x\to c$, the corresponding $t \to c$ as well so $\frac{f(x)}{g(x)} = \frac{f(x)-g(c)}{g(x)-g(c)} \to \lim_{ t \to c } \frac{f'(t)}{g'(t)} $