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Suppose $f(x)$ is continuous on $[a,b)$ and differentiable on $(a,b)$ and that $f '(x)$ tends to a finite limit $L$ as $x \to a^+$. Then $f(x)$ is right-differentiable at $x=a$ and $f '(a)=L$.

(epsilon-delta proof not needed).

This is a practice exam question.

I am having trouble translating this into a 'mathematical' statement. The MVT states that there exists $c$, $a\leq c\leq b$, such that:

$f'(c) = (f(b)-f(a))/(b-a)$

I suppose to prove that $f(x)$ is right differentiable at $x=a$, using the MVT, I need to somehow show that as $x \to a^+$, $f'(c)=f'(a)=L$ ??? Am I on the right track here? Can someone help me get started?

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Apply MVT with $b=x$. As $c$ is sandwiched between $a$ and $x$, then as $x\to a+$ we also get that $c\to ?$. – Jyrki Lahtonen Jun 2 '12 at 11:08

You need to show that $$ \lim\limits_{x\to a+0}\frac{f(x)-f(a)}{x-a}=L $$ By mean value theorem there exist $c(x)\in(a,x)$ such that $$ \frac{f(x)-f(a)}{x-a}=f'(c(x)). $$ Since for all $x\in(a,x)$ we have $c(x)\in(a,x)$, then $\lim\limits_{x\to a+0}c(x)=a$. Thus, $$ \lim\limits_{x\to a+0}\frac{f(x)-f(a)}{x-a}=\lim\limits_{x\to a+0}f'(c(x))=\lim\limits_{t\to a+0}f'(t)=L $$

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thanks for helping me out. Was there any particular reason, why you had c(x)? Rather than just c is a member of (a,x)? Just curious, but this helped a lot! – JackReacher Jun 2 '12 at 11:40
When I write $c(x)$ instead of $c$ I want to emphasize that choice of $c$ depends on $x$. Thus you have a function $c(x)$. – Norbert Jun 2 '12 at 11:43

Let $h$ be a real number such that $h<b-a$. By assumption, $f(x)$ is continuous on $[a,a+h]$ and differentiable on $(a,a+h)$. By mean-value theorem, we have $$\tag{1}f'(c_h)=\frac{f(a+h)-f(a)}{h}$$ for some $c_h\in (a,a+h)$.

By definition, the right derivative of $f$ at $a$ is given by $$\lim_{h\to 0, h>0}\frac{f(a+h)-f(a)}{h}$$ which is equal to $$\lim_{h\to 0, h>0}f'(c_h)$$ by $(1)$. On the other hand, by assumption, $$\lim_{h\to 0, h>0}f'(c_h)=L.$$ The result follows from combining all these.

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Thanks for your help Paul. For (1), should it be h-a? – JackReacher Jun 2 '12 at 12:03

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