Function continuity and differentiability I am given:

$$f(x) =
\begin{cases}
x^{4/3}\sin(1/x)  & \text{if $x\neq0$} \\
0 & \text{if $x=0$}  \\
\end{cases}
$$

I need to show that


*

*$f$ is cts on $\mathbb{R}$

*$f$ is differentiable on $\mathbb{R}$

*$f'(x)$ is unbounded on any open interval containing 0.


Thanks in advnace.
 A: $f$ is  clearly continuous for $x\neq 0$ because $ x^{4/3}\sin(1/x)$ is continuous. Since $$-x^{4/3}\leq x^{4/3}\sin(1/x)\leq x^{4/3}$$ by the squeeze thorem $$\lim_{x\rightarrow 0}x^{4/3}\sin(1/x) = 0$$ so $f$ is also continuous for $x=0$. 
Similarly, $f$ is clearly differentiable at any $x\neq 0$, for $x=0$ use the definition of the derivative to get \begin{eqnarray*}f'(0)& =& \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} \\&=& \lim_{h\rightarrow 0}\frac{f(h)}{h}\\
&=&\lim_{h\rightarrow 0}\frac{h^{4/3}\sin(1/h)}{h}\\
&=&\lim_{h\rightarrow 0} h^{1/3}\sin(1/h)\\
&=&0\end{eqnarray*}
so $f$ is differentiable at $x=0$ and we have 
$$f'(x) = \left\{\begin{array}{cc}4/3x^{1/3}\sin(1/x) + x^{4/3}\cos(1/x)(-\frac{1}{x^2}) & x\neq 0\\
0 & x=0\end{array}\right.$$
Now notice that the term $|x^{4/3}\cos(1/x)(-\frac{1}{x^2})|$ grows unboundedly when $x$ goes to $0$. So the same is true for $f'(x)$.
A: The continuity and differentiability of $f$ on $\mathbb R-\{0\}$ follows immediately. Note, $$\lim_{x\to 0}f(x)=\lim_{x\to 0}\left(x^{4/3}\sin\frac{1}{x}\right)=0=f(0)$$ $$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\left(x^{1/3}\sin\frac{1}{x}\right)=0$$since $\lim_{x\to 0}x^{4/3}=\lim_{x\to 0}x^{1/3}=0$ and $\sin\frac{1}{x}$ is bounded in any deleted neighborhood of $0.$ Consequently $f$ is continuous and differentiable at $0.$ 
