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I'm really struggling with understanding this proof for L'Hopital's rule. So the theorem says that let $-\infty\leq a < b \leq + \infty$, and let $A \in [ -\infty, \infty]$. Assume $f,g :(a,b) \rightarrow \mathbb{R}$ are differentiable on $(a,b)$ and that $g'(x) \neq 0 \forall x \in (a,b)$. If $$\tag{1}\lim_{x\rightarrow a^+} \frac{f'(x)}{g'(x)} = A$$ and either $\lim_{x\rightarrow a^+} f(x) = \lim_{x\rightarrow a^+} g(x) = 0$ or $\lim_{x\rightarrow a^+} g(x) = +/- \infty$, then $\lim_{x\rightarrow a^+} \frac{f(x)}{g(x)} = A$.

Here's the (partial) proof my professor did in class for the case where $A \in \mathbb{R}$:

Let $\epsilon >0$, by (1) we have that $\exists c \in (a,b)$ s.t $\forall x \in (a,c], |\frac{f'(x)}{g'(x)} - A | < \frac{\epsilon}{2}$. Fix $x \in (a, c]$ and let $y<x$. By the Lemma, we have that $\exists t \in (a,c)$ such that $$|\frac{f(x)-f(y)}{g(x)-g(y)} - A| < \frac{\epsilon}{2}.$$ If the $\lim_{x \rightarrow a^x} f(x) = \lim_{x \rightarrow a^x} g(x) = 0$, sending $y$ down to $a$ in (2) yields $$|\frac{f(x)}{g(x)} - A | \leq \frac{\epsilon}{2} < \epsilon.$$

So I'm really confused by the very first statement where we say $\exists c \in (a,b) s.t \forall x \in (a,c]$... Are we using Rolle's theorem? Where did the $x \in (a, c]$ come from? Shouldn't it be $x \in (a,c)$? I'm also guessing that we're using the fact that differentiable functions are continuous to do a $\epsilon$ argument? Insight or a further explanation would be very much appreciated!

Martin Argerami
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Nikitau
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  • http://math.stackexchange.com/questions/1798131/proof-of-lhopitals-rule/1993679#1993679 –  Feb 09 '17 at 21:08

1 Answers1

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It's the definition of limit. Given an arbitrary interval around $A$ (in this case of radius $\varepsilon/2$), there should exist an interval around $a$, $(a,c]$ in this case, such that for every $x$ in $(a,c]$ the value of $f'(x)/g'(x)$ is at less than $\varepsilon/2$ from $A$.

Martin Argerami
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  • Oh wow. I'm feeling really silly now -- you're right. So this would be the definition where "a sequence converges if $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ s.t $\forall n > N, d(p_n, p) < \epsilon$. – Nikitau Dec 07 '16 at 04:01
  • More properly, it would be the limit of a function, where $\lim_{x\to y}f (x)=A $ means that for every $\varepsilon >0$ there exists $\delta>0$ such that whenever $|x-y|<\delta $, we have $|f (x)-A|<\varepsilon $. – Martin Argerami Dec 07 '16 at 09:38