Rudin Chp. 5, q13: Proving that $f'(0)$ exists only if $a > 1$ 
Rudin Chp. 5 q. 13:
Suppose $a$ and $c$ are real numbers, $c > 0$, and $f$ is defined on $[-1, 1]$ by
$$f(x) = x^a \sin(|x|^{-c}), x≠0$$
$$f(x) = 0, x=0$$
(b) $f'(0)$ exists iff $a > 1$

To me, it seems quite clear that $a>1$ would work because it is intuitively clear that $f(x) → 0$ as $x → 0$. The function $\sin(u)$ has a range of $[-1, 1]$, so while $\sin(|x|^{-c})$ will oscillate infinitely as $x→0$, $x^a → 0$ for $a > 0$. It is clear that this is continuous for $a>0$.
But I need to show that it is differentiable for $x=0$ iff $a>1$. And this is where I have gotten stuck. I am able to show that it is not differentiable for $a ≤ 1$. But when I try to show it is differentiable for $a>1$, I fail to do so. I tried to differentiate $f(x)$ in general ($f'(x)$) then show it will not work as $x→0$ for $a≤1$, but this method does not work with $a>1$, and I end up with $x^{a+1} / |x|^{-c-2}$ (plus unimportant constants and cosine). And that is bad because, for example, if a = 2 and c = 10, that limit clearly diverges to infinity.
A fellow student claimed to used the definition of the derivative to solve this and I tried this:
$$f'(x) = \lim_{t→x} \frac{f(t) - f(x)}{t-x} = \lim_{t→x} \frac{t^a \sin|t|^{-c} - x^a \sin|x|^{-c}}{t-x}$$
And we are interested in only $f'(0)$, so we can simply:
$$f'(0) = \lim_{t→0} \frac{f(t) - f(0)}{t-0} = \lim_{t→0} \frac{t^a \sin|t|^{-c} - 0}{t}= \lim_{t→0} t^{a-1} \sin|t|^{-c}$$
Assume $a>1$
$$=\left[\lim_{t→0} t^{a-1}\right]\left[\lim_{t→0}\sin |t|^{-c}\right] = 0\left[\lim_{t→0}\sin|t|^{-c}\right] = 0$$
This is clear because $\sin(u)$ has a range of [-1, 1].
Clearly in the case that $a≤1$, this will diverge.
Is this all that I need to do? I don't understand why my first method did not work but the second did, if that is indeed all I must do.
I am worried about the main concept, not about how my “proof” looks. I can write it out MUCH better on paper, I am struggling to format this well on the computer (and sorry for this!)

From $\lim_{t\to0}t^{a-1}\sin(|t|^{-c})$, I would use the Squeeze Theorem, since this is between $-|t|^{a-1}$ and $|t|^{a-1}$.  You have written the limit of the sine, which of course doesn't exist.  
