# Derivative of functions at a point

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?

• Does $\lim\limits_{x\to 0}\dfrac{\sin x}{x^2}$ exist? – Anupam Feb 5 '14 at 14:50
• I think it doesn't. So $f$ is not differentiable at $x=0$. – Anupam Feb 5 '14 at 14:51
• but its $\frac{\sin x}{x}$ in the question. – Suraj M S Feb 5 '14 at 14:52
• $f^{'}(0)=\lim\limits_{x\to 0}\dfrac{\sin x}{x^2}$. Isn't it? – Anupam Feb 5 '14 at 14:56
• 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 $x=0$. – 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.

$f(x)=\frac{sinx}{x}$

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

R.H.D

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

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

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.

THANKS.

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