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?
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.