If the function $f:[0,1]\to \mathbb R$ defined by $f(x)=\dfrac{\sin x}{x}$ when $x\neq 0$ and $f(0)=0$, then is $f$ differentiable at $x=0?$

I think by using L'Hospital's rule $f$ is differentiable more than one time. Am I right?

  • $\begingroup$ Does $\lim\limits_{x\to 0}\dfrac{\sin x}{x^2}$ exist? $\endgroup$ – Anupam Feb 5 '14 at 14:50
  • $\begingroup$ I think it doesn't. So $f$ is not differentiable at $x=0$. $\endgroup$ – Anupam Feb 5 '14 at 14:51
  • $\begingroup$ but its $\frac{\sin x}{x}$ in the question. $\endgroup$ – Suraj M S Feb 5 '14 at 14:52
  • 1
    $\begingroup$ $f^{'}(0)=\lim\limits_{x\to 0}\dfrac{\sin x}{x^2}$. Isn't it? $\endgroup$ – Anupam Feb 5 '14 at 14:56
  • $\begingroup$ Sorry about my previous comments (I didn't notice you defined $f(0)=0$; did you want $f(0)=1$?). Yes, that's right. $f$ is not differentiable at $x=0$. Note $f$ isn't even continuous at $x=0$. $\endgroup$ – David Mitra Feb 5 '14 at 15:03

You probably meant $f(0):=1$. In this case $f$ is differentiable for the following reason. The function $\sin x$ can be developed into a power series

$$ \sin x = \sum_{n=0}^\infty \frac{x\cdot(-x^2)^n}{(2n+1)!}\,. $$

Deviding by $x$ gives you:

$$ f(x) = \sum_{n=0}^\infty \frac{(-x^2)^n}{(2n+1)!}\,. $$

This power series still converges for all $x\in\mathbb R$. A power series is always differentiable by differetiating term by term. It is even differentiable arbitrarily often. You can see that $f$ is an even function. Therefore $f'(0)=0$.


As $$\lim_{x\to 0}f(x)=\lim_{x\to 0}\dfrac{\sin x}{x}=1\ne0=f(0),$$ $f$ is not even continuous in 0, so can't be differentiable.



To check differentiability at $x=0$ you can use the fundamental theorem of differentiability


$f'(x)=limh \to 0\frac {f(x+h)-f(x)}{h}$

$f'(x)=limh \to 0\frac {\frac{sin(x+h}{x+h}-\frac{sinx}{x}}{h}$

on expanding sin series u get the value of R.H.D=1

again similarly L.H.D=1

since both the values are equal but unequal to the function value @ x=0 hence$f(x)$ is not differentiable at $x=0$.I hope now u find the general way.


  • $\begingroup$ This is not the way to find the derivative of a function $\endgroup$ – Anupam Feb 6 '14 at 2:18
  • $\begingroup$ @Anupam why? what's wrong with this? $\endgroup$ – Danny obama Feb 7 '14 at 16:52
  • $\begingroup$ yeah it was right...I couldn't follow properly. Sorry... $\endgroup$ – Anupam Feb 7 '14 at 17:46
  • $\begingroup$ @Anupam so there is no need to down vote me :-) $\endgroup$ – Danny obama Feb 7 '14 at 17:56
  • $\begingroup$ yeah I tried to upvote again but it is locked. Please edit the answer a bit. $\endgroup$ – Anupam Feb 7 '14 at 17:58

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