Which of the following about $f(x)$ is correct? Let $f:\mathbb R \to \mathbb R$ be defined by 
$$f(x) =     \left\{
\begin{array}{1l}
      \dfrac{\sin x}{x} &amp; x \neq 0 \\
      1 &amp; x = 0 \\
\end{array} 
\right.$$
Then:


$f$ is not continuous.
$f$ is continuous but not differentiable.
$f$ is differentiable



  Solution:
  Clearly $f(x)$ is continuous at $x=0$ and $\lim_{x\to 0^{+}}\frac{\sin x}{x}=1=\lim_{x\to 0^{-}}\frac{\sin x}{x}$.


Now, about differentiability at $x=0$,
$\lim_{x\to 0^{+}}\frac{\frac{\sin h}{h}-1}{h}=\frac{\sin h}{h^2}-\frac{1}{h}$ and $\lim_{x\to 0^{-}}\frac{1-\frac{\sin (-h)}{-h}}{h}=-\frac{\sin h}{h^2}+\frac{1}{h}$ .
Now both these limits should go to $0$ as $f'(0)=0$. But they are not going to zero. Hence, the function is continuous but not differentiable.
:Option (3) is correct

$\lim_{h\to 0^{+}}\frac{\frac{\sin h}{h}-1}{h}$ $(\frac{0}{0}) $ form.
Now use L'Hospital's rule we get 
$\lim_{h \to 0^{+}}\frac{\frac{h\cos h-\sin h}{h^2}}{1}
 =\lim_{h \to 0^{+}}\frac{h\cos h-\sin h}{h^2}(\frac{0}{0})form$ 

again using L'Hospital's rule,
$\lim_{h \to 0^+}(-\frac{sinh}{2})=0$

Similarly using L'Hospital's rule $ \lim_{h\to 0^{-}}\frac{\frac{\sin h}{h}-1}{h}(\frac{0}{0} form)=0$
Thus $$\lim_{h\to 0}\frac{\frac{\sin h}{h}-1}{h}=0$$
Thus when $x\ne 0,f'(x)=\frac{x\cos {x}-\sin {x}}{x^2}$ 
And when $x=0 ,f'(x)=0$
Actually $f$ is infinitely differentiable

Your limit is right $ \dfrac{\sin(h)}{h^2}=\frac{h-h^3/6}{h^2}+o(h) $ developing $\sin$ asyptotically at second order so $\dfrac{\sin(h)}{h^2}-\dfrac{1}{h}=-h/6+o(h)$ Actually you're dealing with $sinc$ (say cardinal sinus function). $f$ would by infinitely differentiable. In order to show it you can develop it into  a serie.

$$ \forall x\in \mathbb{R}\ , \  f(x)=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{(2n+1)!}$$

so  $f$ is infinitely differentiable on its circle of convergence  (here $\mathbb{R}$  because the radius of convergence equal to $\infty$)