Look carefully... the definition $\displaystyle x^m \sin \frac1{x}$ holds only for $\displaystyle x \neq 0$. You cant simply substitute 0 in the place of x to check for continuity.
Thus your problem for m = 0 reads:
$\displaystyle f(x) = \begin{cases} \sin \frac1{x} & x\neq 0 \\ 0 & x = 0 \end{cases}$
Is f(x) continuous at x = 0?
(This is a very popular problem. I am sure you would have seen it somewhere in your notes/text)
The function is well defined. The point of my post was to ask if f(x) is continuous at 0. I claim it is not. Can you prove it?
To prove it is discontinuous at 0, observe that sequence $\displaystyle x_n = \frac{2}{(2n+1)\pi}$ converges to 0, but $\displaystyle f(x_n) = (-1)^n$
First prove the following result:
(*)$\displaystyle \lim_{x \to 0} x^m \sin \frac1{x}$ exists and equal to 0 iff $\displaystyle m > 0$.
Now for part (A) we need to find m for which $\displaystyle \lim_{x \to 0} f(x) = f(0) = 0$, Thus we need m for which $\displaystyle \lim_{x \to 0} x^m \sin \frac1{x} = 0$. From the result (*), this happens iff $\displaystyle m > 0$.
For part (B), we need values of 'm' for which $\displaystyle \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0}$ exists. But $\displaystyle \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0} = x^{m-1} \sin \frac1{x}$. Again by the result, the limit exists iff $\displaystyle m-1 > 0 \implies m > 1$
Do part (C) yourself, its instructive
No! A function is continuously differentiable at x = 0 means the derivative at 0 exists and the derivative is continuous at 0 ! I never proved any result about the derivative!
Ok I will do the last one too... But I will let you prove some statements on your own.
Since we have already proved that m > 1 iff f(x) is differentiable at x = 0, for continuous differentiability we need m > 1.
$\displaystyle f'(x) = x^{m-2} \left(mx \sin \frac1{x} - \cos \frac1{x}\right)$
Now first prove that if m > 2, then the function f'(x) is continuous at 0, using (*) and making a similar rule for $\displaystyle \lim_{x \to 0} x^m \cos \frac1{x}$(prove this too!).
Now the only remaining case is m=2. For this case,$\displaystyle f'(x) = \begin{cases} 2x\sin \frac1{x} - \cos \frac1{x} & x \neq 0 \\ 0 & x = 0 \end{cases}$.
Choose the sequence $\displaystyle x_n = \frac1{n \pi}$, and observe that $\displaystyle x_n \to 0$ but $\displaystyle f(x_n) = -(-1)^n$. Thus f(x) is not continuously differentiable at x = 0.
So at the end, let me list the exercises you have to do, to complete the proof rigorously:
(*) exists and equal to 0 iff $\displaystyle m > 0$
(**) $\displaystyle f'(x) = x^{m-2} \left(mx \sin \frac1{x} - \cos \frac1{x}\right)$
(***) $\displaystyle \lim_{x \to 0} x^m \cos \frac1{x}$ exists and equal to 0 iff $\displaystyle m > 0$
(****) Using (**) and (***), prove that if m > 2, then $\displaystyle f'(x)$ is continuous at $\displaystyle x = 0$
That solution you got is of no use to part A. To demonstrate discontinuity at $\displaystyle x_0$, we have to find a sequence $\displaystyle x_n \to x_0$, but the sequence $\displaystyle f(x_n) \not \to f(x_0)$.
Your sequence $\displaystyle x_n \to 0$, but $\displaystyle f(x_n) = \sin 2n \pi = 0$. So it is of no use.
As a side note, I am really sorry transgalactic, that I am failing in explaining this concept to you. I have already given you the answer for this in my previous posts. The sequence you can use is $\displaystyle x_n = \frac2{(2n+1)\pi}$
I am telling the following based on our discussion.....If you take it in the right spirit, I suggest you take tutions. Sometimes a live tutor teaching a new concept is an invaluable experience.