Let $f$ be a real valued continuous function defined on $[0,2]$ such that $f$ is differentiable at all point except possibly at $1$. Suppose that $\lim_{x\to 1}f^{'}(x)=5.$

Show that $f$ is differentiable.

In order to show that $f $ is differentiable we have to show its differentiability only at $x=1$.

So we have to show that $\lim_{x\to 1}\dfrac{f(x)-f(1)}{x-1}$ exists. Since $\lim_{x\to 1}f^{'}(x)=5.\implies $ Given $\epsilon >0$ we have $\delta>0$ such that $|f^{'}(x)-5|<\epsilon $ whenever $0<|x-1|<\delta$

But $f^{'}(1)=\lim_{x\to 1}\dfrac{f(x)-f(1)}{x-1}$.

I am unable to proceed after that.Please give some hints to start.


Using MVT: $f(x) - f(1) = f'(c)(x-1)$. Can you continue?

  • $\begingroup$ As $f$ is continuous on $[0,1]$ and differentiable in $(0,1)$ so applying MVT on $f(x)$ in $[0,1]$ $\exists c\in (0,1)$ such that $f(x)-f(1)=f^{'}(c)(x-1)\implies \lim_{x\to 1^-}\dfrac{f(1)-f(x)}{1-x}=f^{'}(c)$ again applying MVT on $(1,2)$ we have $\lim_{x\to 1^+}\dfrac{f(x)-f(1)}{x-1}=f^{'}(d)$ where $d\in (1,2)$ $\endgroup$ – Learnmore Mar 17 '16 at 10:24
  • $\begingroup$ But how to show that the above two limits are equal ;do we require $\lim f^{'}(x)=5$ $\endgroup$ – Learnmore Mar 17 '16 at 10:25
  • $\begingroup$ Can you please explain here what to do next? $\endgroup$ – Learnmore Mar 17 '16 at 10:26
  • $\begingroup$ For $0<x<2$ and$ x\ne 1$ we have $f(x)-f(1)=(x-1)f"(c_x) $ where $c_x$ lies between $x$ and $1$. We know this without knowing in advance whether $f'(1)$ exists: It is sufficient that $f(y$) is continuous at $y=1$ and is differentiable on $(0,1)\cup (1,2).$ Now as $x\to 1$ we have $ c_x\to 1$ so $f'(c_x)\to 5.$ Hence $\lim_{x\to 1}(f(x)-f(1)/(x-1)=\lim_{x\to 1}f'(c_x)=5.$ $\endgroup$ – DanielWainfleet Mar 17 '16 at 11:05

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