Solution: It is clear that the function is continuous and all derivative exists other than $0$. We need to check the continuity and differentiability of the given function at $x=0$.

---

Let us start with the continuity of the function at $0$. Note that \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin x}{x} = 1 = f(0). \] Therefore, the function is continuous at $x=0$.

---

Now we will discuss the differentiability of the function. One easy way to look at this problem is to consider the Taylor series of the $\sin $ function. We have \begin{align*} & \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \\ \implies & \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{4!} + \cdots. \end{align*} Thus the function is differentiable $0$. Moreover, this also implies all the derivatives of the given function exists at $x=0$.

---

---

In order to see the same with the definition of differentiation, we consider the following limit. \begin{align*} \lim_{h \to 0} \frac{f(0+h)-f(0)}{h} & = \lim_{h \to 0} \frac{\frac{\sin h}{h}-1}{h} \\ & = \lim_{h \to 0} \frac{\sin h - h}{h^2}. \end{align*}
As the above limit is in the form of $\frac{0}{0}$, we will apply the L'Hospital rule. So, \begin{align*} \lim_{h \to 0} \frac{\sin h - h}{h^2} & = \lim_{h \to 0} \frac{(\sin h-h)^\prime }{\left( h^2 \right)^\prime } \\ & = \lim_{h \to 0} \frac{\cos h-1}{2h} \\ & = \lim_{h \to 0} \frac{(\cos h-1)^\prime }{(2h)^\prime } \\ & = \lim_{h \to 0} \frac{-\sin h}{2} \\ & = 0. \end{align*} Thus, the function is differentiable at $x=0$. The derivative will be \[ f^\prime (x) = \begin{dcases} \frac{x \cos x-\sin x}{x^2}, &\text{ if } x\neq 0 ;\\ 0, &\text{ if } x=0. \end{dcases} \]

---

Now let us see whether the derivative function is continuous or not. For that, consider the following limit and will apply the L'Hospital rule various times. \begin{align*} \lim_{x \to 0} f^\prime (x) & = \lim_{x \to 0} \frac{x\cos x -\sin x}{x^2} \\ & = \lim_{x \to 0} \frac{\cos x-x\sin x-\cos x}{2x} \\ & = \lim_{x \to 0} \frac{-\sin x}{2} \\ & = 0. \end{align*} Thus, the derivative function is continuous.