Show the function's $f(x)=\begin{cases}x^m\sin\frac{1}{x}&\text{if} \ x\ne0 \\ 0 &\text{if} \ x=0 \end{cases}$differentiability at following points 
Show that the function 
  $$f(x) = 
\begin{cases}
x^m \sin \frac{1}{x} & \text{if} \ x\ne0 \\ \\
0 &\text{if} \ x=0 \ 
\end{cases}.
$$
  is:$(i)$ differentiable at $x=0$, if $m \gt 1$
  $(ii)$ continuous but not differentiable at $x=0$, if $0 \lt m \lt 1$
  $(iii)$ neither continuous nor differentiable, if $m \le 0$

I tried to solve it like this:
$$\lim_{x\to 0^-}f(x)=\lim_{h\to0}f(0-h)=\lim_{h\to0}(-h)^m\sin\frac{1}{-h}=0$$and
$$\lim_{x\to 0^+}f(x)=\lim_{h\to0}f(0+h)=\lim_{h\to0}(h)^m\sin\frac{1}{h}=0$$
for every $x\gt0$(Why???)
The other two I was not able to find a proper answer.
Thankyou
 A: Disclaimer
Since not all powers of negative numbers are defined, you may add some more restrictions to domains of either $m$ or $x$. Let's restrict $x \ge 0$, so in this case you can analyse the whole range of powers $x^m$, both positive and negative reals, as it's stated in your post. But you should keep in mind that both continuity and differentiability will be only one-sided.
Continuity
Your calculations are quite correct to justify that the function
$$
f(x) = \left \{ \begin{array}{ll}
x^m \sin \frac 1x & x > 0 \\
0 & x = 0
\end{array}\right .
$$
is indeed right-side continuous at $0$ when $m > 0$. It follows from the fact that
$$
\lim_{x \rightarrow 0 + 0} x^m = 0
$$
for $\forall m>0$ and $\sin \frac 1x$ is bounded as $-1 \le \sin \frac 1x \le 1$ for $\forall x$, therefore their product is also approaches $0$ which is $f(0)$, so it's continuous from the right side.
If $m = 0$, then for $\forall x \ge 0$, your function will be
$$
f(x) = \left \{ \begin{array}{ll}
\sin \frac 1x & x > 0 \\
0 & x = 0
\end{array}\right .
$$
Due to the periodical nature of sine function, for any given open interval containing $0$ it will be taking any value in $[-1,1]$, so it doesn't have limit, hence not continuous at $0$.
If $m < 0$, you can rewrite $f(x)$ as
$$
f(x) = \left \{ \begin{array}{ll}
\left(\frac 1x\right)^{-m} \sin \frac 1x & x > 0 \\
0 & x = 0
\end{array}\right .
$$
First term gets infinitely large when $x \rightarrow 0+0$ and sine is still bounded but also changes its sign, so their product doesn't have limit at all. So, again, it's not continuous at $0$.
Differentiability
If you use the very basic definition of differentiability, then you can get
$$
f'(0) = \lim_{x \rightarrow 0+0} \frac {f(x)-f(0)}x = \lim_{x \rightarrow 0+0} x^{m-1} \sin \frac 1x
$$
So it's pretty much the same, except order is $m-1$ now. Using conclusions of previous calculations, you may state that if $m-1>0$ or $m>1$ function $f(x)$ is right-side differentiable and continuous at $0$, if $0 \le m \le 1$ is not right-side differentiable but right-side continuous at $0$ and finally if $m \le 0$ is neither right-side differentiable not right-side continuous at $0$.
