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Define $h(x) = 1$ except at $1$ where $h(1) = 0$. Also define $H(x) = \int_0^x h(t)$. Now I tried to show that $H$ is differentiable at $1$. My proof is to compute $$ \lim_{x \to 1^-} {H(1) - H(x)\over 1-x} = 1$$ and $$ \lim_{x \to 1^+} {H(x) - H(1)\over x-1} = 1$$

Hence $H$ is differentiable at $1$ and $H'(1) = 1$. Is it correct?

blue
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    You can get $H(x)$ explicitly and the result follows. How have you computed the limits? – mfl Jul 10 '14 at 12:47
  • @mfl I computed the limits like this: $$ \lim_{x\to 1^-} {H(x)-H(1)\over x-1} = \lim_{x\to 1^-}{\int_0^x 1 - \int_0^1 1 \over x-1} \stackrel{de l'Hopital}{=} \lim_{x \to 1^-} {x-0 \over 1}=1$$ – blue Jul 11 '14 at 09:07

1 Answers1

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The question reads as: "I used the definition of derivative to prove that $H'(1)=1$. Therefore, $H'(1)=1$. Is this correct?"

Yes, it's correct. If you can prove that $H'(1)=1$, then it follows that $H'(1)=1$.

As mfl noted, $H(x)=x$ for all $x$.