Differentiability of $f(x)=sin(x)/x$ if $x\ne0$ and $1$ if $x=0$ I am trying to see if 
$$f(x)= \begin{cases} \frac{\sin(x)}x &\text{ if x}\neq0\\ 1 &\text{ if x}=0. \end{cases} $$
is differentiable more than once.
This is what I did:
$$f'(0)= \begin{cases} 0 \\ 0 . \end{cases} $$
$$f''(0)= \begin{cases} -1/3 \\ 0. \end{cases} $$
Hence it is differentiable only once.
Also, $$g(x)= \frac{\sin(x)}x$$ is differentiable more than once.
Am I right?
 A: I don't think so.
Based on the definition of $f(x)$, we can get the expression of its first derivative
$$
f'(x)=\begin{cases}0,&\text{if}\quad x=0\\\frac{x\cos(x)-\sin(x)}{x^2}, &\text{if}\quad x\neq0\end{cases}
$$ 
Then 
$$f''(0)=\lim_{x\rightarrow0}\frac{f'(x)-f'(0)}{x-0}=\lim_{x\rightarrow0}\frac{x\cos(x)-\sin(x)}{x^3}=-\frac{1}{3}$$
and
$$
f''(x)=\frac{2\sin(x)-2x\cos(x)}{x^3}-\frac{\sin(x)}{x}\quad\text{if}\quad x\neq 0
$$
As
$$
\lim_{x\rightarrow0}f''(x)=-\frac{1}{3}=f''(0)
$$
we can differentiate it more than twice.
A: This function $f$, called "sinc" in signal processing, is infinitely differentiable at $0$. This becomes immediately clear when we write $f$ in the form
$$f(x)=\int_0^1\cos(t x)\>dt$$
(check that this is valid for all $x\in{\mathbb R}$). You could also invoke the series representation
$$f(x)=1-{x^2\over 3!}+{x^4\over 5!}-{x^6\over 7!}+\ldots$$
coming from the sin series.
PS:  Of course it's a useful exercice to prove the differentiability of $f$ at $0$ "by hand".
A: For $x\ne0$, $f'(x)=\dfrac{x\cos x-\sin x}{x^2}$ and $f$ is differentiable.
For $x=0$, by the definition of the derivative
$$f'(x)=\lim_{x\to0}\frac{\frac{\sin x}{x}-1}x=\lim_{x\to0}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}\cdots-x}{x^2}=0.$$
(It coincides with $\lim_{x\to0}f'(x)$ as found above, hence the first derivative is continuous.)
The second derivative at $x=0$ is
$$\lim_{x\to0}\frac{\frac{x\cos x-\sin x}{x^2}-0}x=\lim_{x\to0}\frac{x-\frac{x^3}{2!}+\frac{x^5}{4!}\cdots-x+\frac{x^3}{3!}-\frac{x^5}{5!}\cdots}{x^3}=-\frac1{3}.$$
