# Proving L'Hospital's theorem using the Generalized Mean Value Theorem

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?

• For $x \neq c$, you have $$\frac{f(x)}{g(x)} = \frac{f(x) - f(c)}{g(x) - g(c)}.$$ Now apply the GMVT with $a = x$, $b = c$, and call the $c$ in the statement of the GMVT (which has nothing to do with the $c$ in the statement of L'H) for example $y$ (or $\xi$). – Daniel Fischer Oct 19 '19 at 13:45
• @DanielFischer: I am still not clear about it. If I take limit as $x\to c$ (and has to be $c$ because that is where $f$ and $g$ are $0$), then I don't see any way to get the L'Hospital's rule. It would basically say $$\lim\limits_{x\to c}\frac{f^{\prime}(\xi)}{f^{\prime}(\xi)} = L$$ – Peaceful Oct 19 '19 at 14:16

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 $$\lim_{x \to c} \frac{f(x)}{g(x)} = L.$$