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This is an exercise I have been working on today from Stephen Abbott's Understanding Analysis:

Exercise 5.3.11. Use the Generalized Mean Value Theorem to furnish a proof of the $0/0$ case of L'Hospital's rule (Theorem 5.3.6.).

Recall the $0/0$ case of L'Hospital's rule:

Theorem 5.3.6. (L'Hospital's rule: $0/0$ case). Assume $f$ and $g$ are continuous functions defined on an interval containing $a$, and assume that $f$ and $g$ are differentiable on this interval, with the possible exception of the point $a$. If $f(a)=0$ and $g(a)=0$, then \lim_{x\to a}\frac{f'(x)}{g'(x)}=L\Longrightarrow\lim_{x\to a}\frac{f(x)}{g(x)}=L.

Also recall the Generalized Mean Value Theorem:

Theorem 5.3.5. (Generalized Mean Value Theorem). 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'(c)=[g(b)-g(a)]f'(c). If g' is never zero on $(a,b)$, then the conclusion can be stated as \frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.

First of all, I am a little confused about the theorem's statement: is it asking me to assume that both $f(a)=g(a)=0$ and \lim_{x\to a}f'(x)/g'(x)=L to show that $\lim_{x\to a}f(x)/g(x)=L$? If so, can I introduce a subinterval $(a,b_1)$ such that \frac{f'(a)}{g'(a)}=\frac{f(b_1)-f(a)}{g(b_1)-g(a)}=\frac{f(b_1)}{g(b_1)}? I am not sure, since the theorem states that $f$ and $g$ may not necessarily be differentiable at $a$. Nevertheless, I have the feeling that I could be close if it were not for my shortsightedness. -_-

But anyway, what do you guys think? And thanks in advance!

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    @joriki I am sorry as I did $n$ot see that! And, that is also standard to assume in L'Hospital rule. I am deleting that comment!2012-03-01

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