The following is the theorem from Understanding Analysis by Stephen Abbott.
Thm 5.3.6 : (L'Hospital's rule: $0/0$ case) Let $f$ and $g$ be continuous on an interval containing $a$, and assume $f$ and $g$ are differentiable on this interval with the possible exception of the point $c$. If $f(c)=g(c)=0$ and $g^{\prime}(x)\neq 0$ for all $x\neq c$, then
$$\lim\limits_{x\to c}\frac{f^{\prime}(x)}{g^{\prime}(x)} = L \implies \lim\limits_{x\to c}\frac{f(x)}{g(x)} = L$$
In the proof, the author says that "The argument follows from a straightforward application of the generalized mean value theorem". However, I am failing to see any such "straightforward" argument.
As given in the same book, the Generalized Mean value Theorem (GMVT), if $f$ and $g$ are continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists a point $c\in (a,b)$ where
$$[f(b)-f(a)]g^{\prime}(c)=[g(b)-g(a)]f^{\prime}(c)$$
To use it to prove the L'Hospital's rule, one must first try to see whether a closed interval $[a, b]$ around $c$ could be constructed so that $c$ will satisfy the mean value theorem. However, I think that the converse doesn't exist.Even if it did, it is not clear to me how would one proceed to prove the required theorem. Could someone please help me?
Best Answer
Given $\varepsilon > 0$ we want to show the existence of a $\delta > 0$ such that $$\biggl\lvert \frac{f(x)}{g(x)} - L \biggr\rvert < \varepsilon$$ for $0 < \lvert x - c\rvert < \delta$ (and $x$ belonging to the interval). By the assumption $$\lim_{x \to c} \frac{f'(x)}{g'(x)} = L$$ there is a $\delta > 0$ such that $$\biggl\lvert \frac{f'(y)}{g'(y)} - L \biggr\rvert < \varepsilon$$ for all $y$ in the interval with $0 < \lvert y - c\rvert < \delta$. The generalised mean value theorem shows that this same $\delta$ also works for $f(x)/g(x)$, since for $0 < \lvert x - c\rvert < \delta$ we have \begin{align} \biggl\lvert \frac{f(x)}{g(x)} - L\biggr\rvert &= \biggl\lvert \frac{f(x) - f(c)}{g(x) - g(c)} - L\biggr\rvert \\ &= \biggl\lvert \frac{f'(y)}{g'(y)} - L \biggr\rvert \\ &< \varepsilon \end{align} for some $y \in (c,x)$ or $y \in (x,c)$, depending on whether $x > c$ or $x < c$.
Thus, since $\varepsilon > 0$ was arbitrary, it follows that $$\lim_{x \to c} \frac{f(x)}{g(x)} = L.$$