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Let $ f $ be a function that is differentiable on a deleted neighborhood of $x_0 ∈ R$ and continuous at $x_0$. Show that if $\lim \limits_{x \to x_0} f'(x) = L$ then $f'(x_0) = L $

Bach
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Rizon
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2 Answers2

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@Rizon,

You can apply the MVT on $[x_0,x_0+h]$ with $h>0$ to say that $$ \frac{f(x_0+h)-f(x_0)}{h}=f'(c) $$ for some $c\in(x_0,x_0+h)$. By taking the limit as $h\to 0^+$, and using the fact that $\lim\limits_{x\to x_0} f'(x)=L$, the definition of (right) derivative and the fact that $c\to x_0$ as $h\to 0^+$, you conclude that $$ f'_+(x_0)=L $$ You can do the same for the left derivative, which will also be $L$. Observe that we use the continuity at $x_0$ to be able to apply MVT on both sides. Hope this helps.

GReyes
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The L'Hopital's rule solution. We want to compute, $$ \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} $$ which is of indeterminate form $0/0$. So, applying L'Hopital's and noting that the rule only requires differentiability in a deleted neighborhood of $x_0$, we have $$ \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}= \lim_{x\to x_0}\frac{f'(x)}{1}=L $$ by assumption.

operatorerror
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    @ qbert L'Hopital's rule needs your function to have a limit at the given point. What I did was essentially the proof of L'Hopital's rule. – GReyes Jan 14 '19 at 02:23
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    @GReyes, $f$ is continuous at $x_0$, so it has a limit at that point. So using L'Hopital's is great. – Rizon Jan 14 '19 at 02:40
  • @GReyes yes of course, that is why we get the indeterminate form. Also, yes you did essentially the proof of L'Hopital's rule – operatorerror Jan 14 '19 at 02:50