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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?

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  • $\begingroup$ 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 $\xi$). $\endgroup$ – Daniel Fischer Oct 19 '19 at 13:45
  • $\begingroup$ @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 $$\lim\limits_{x\to c}\frac{f^{\prime}(\xi)}{f^{\prime}(\xi)} = L$$ $\endgroup$ – Peaceful Oct 19 '19 at 14:16
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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 $$\lim_{x \to c} \frac{f(x)}{g(x)} = L.$$

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