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Indeed, my question is very similar to Is the right-hand derivative equal to the right-hand limit of the derivative?, but with different conditions (there is no assumption about the differentiability).

The hypothesis is: $\lim_{x \to a^+}f'(x)=f'_+(a)$ if the expresions on both of the sides exist. (where $f'_+(a)=\lim_{h \to 0^+} \frac{f(a+h)-f(a)}{h}$)

I have seen it in my analysis book but am not able to give a rigorous proof. So far, I have tried to use the definitions to derive $|\lim_{x \to a^+}f'(x)-f'_+(x)|<\epsilon$ for all $\epsilon>0$, but with no result. Does anyone know a better way?

My another hypothesis is that if $\lim_{x \to a^+}f'(x)$ exists and $f$ is continuous at $a$ then the above mentioned equality holds.

May I ask you for help?

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  • $\begingroup$ Did you mean $f'_+(a)$? $\endgroup$ – copper.hat Feb 3 '17 at 21:33
  • $\begingroup$ Oh, my fault. Yes, I meant. $\endgroup$ – Greenhorn3.14 Feb 3 '17 at 21:35
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    $\begingroup$ Use the MVT on the difference quotient. $\endgroup$ – zhw. Feb 3 '17 at 21:39
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    $\begingroup$ @zhw.: You should make that an answer. $\endgroup$ – copper.hat Feb 3 '17 at 21:40
  • $\begingroup$ OK I will do that. $\endgroup$ – zhw. Feb 4 '17 at 21:18
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For small $h>0,$

$$\tag 1\frac{f(a+h)-f(a)}{h} = f'(c_h)$$

for some $c_h\in (a,a+h),$ by the mean value theorem (make sure to verify the MVT can be applied here). As $h\to 0^+,$ $ c_h \to a^+.$ Since we are given $\lim_{x\to a^+} f'(x) = L$ for some $L,$ we conclude $f'(c_h)\to L.$ By $(1)$ the right-hand derivative of $f$ at $a$ is $L.$

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