Continuity And Derivative Of A Function 
Let $
f(x)=
\begin{cases}
\sin^2x \sin(\frac{1}{x}), x\neq 0\\
0, x=0\\
\end{cases}
$ 
a. Find all the points where $f(x)$ is continuous
b. Find $f'(x)$ for all the points $f(x)$ is continuous
c. Is $f'(x)$ continuous where it is defined? if not where and which discontinuity type is it?

So for a we look at $\sin^2x \sin(\frac{1}{x})$ at $x\neq 0$ this is a product of two continuous functions and therefore continuous for $x=0$ we look at $\lim_{x\to 0}\sin^2x \sin(\frac{1}{x})\leq \lim_{x\to 0}|\sin^2x \sin(\frac{1}{x})|=\lim_{x\to 0}\sin^2x |\sin(\frac{1}{x})|\leq 0\cdot 1=0$   so $$
f(x)=
\begin{cases}
\sin^2x \sin(\frac{1}{x}), x\neq 0\\
0, x=0\\
\end{cases}
$$  is continuous for all $x$
b. For $x\neq 0$ we can take $$[\sin^2x \sin(\frac{1}{x})]'=2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$$ (not sure why can we do it)
At $x=0$ we look at $$lim_{\Delta h\to 0}\frac{\sin^2(x+\Delta h) \sin(\frac{1}{x+\Delta h})-\sin^2x \sin(\frac{1}{x})}{\Delta h}\mid_{x=0}=lim_{\Delta h\to 0}\frac{\sin^2(\Delta h) \sin(\frac{1}{\Delta h})-0}{\Delta h}=lim_{\Delta h\to 0}\frac{\sin(\Delta h)}{\Delta h} \cdot sin(\Delta h)\cdot\sin(\frac{1}{\Delta h})\leq lim_{\Delta h\to 0}\mid\frac{\sin(\Delta h)}{\Delta h} \cdot sin(\Delta h)\cdot\sin(\frac{1}{\Delta h})\mid=lim_{\Delta h\to 0}\mid\frac{\sin(\Delta h)}{\Delta h}\mid \cdot \mid \sin(\Delta h)\mid \cdot \mid\sin(\frac{1}{\Delta h})\mid=1\cdot 0 \cdot 1=0$$
(When can I take the limit of a product of functions? just when I am sure the the limit of each function is finite?)
c. Now we have to check that $\lim_{x\to 0}2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$
(how do I find if this limit exist?) 
In general are a and b are valid?
 A: First, the product of continuously differentiable maps is continuously diufferentiable. As $x \mapsto \sin^2 x$ and $x \mapsto \sin \left( \frac{1}{x} \right)$ are both continuously differentiable for $x \neq 0$, $f$ is continuously differentiable for all $x \neq 0$.
Also, for all $x \in \mathbb R$, you have $0 \le \vert x \vert \le \vert \sin x \vert$ and therefore
$$0 \le \vert f(x) \vert = \left\vert \sin^2x \sin \left(\frac{1}{x}\right) \right\vert \le x^2 \left\vert \sin \left(\frac{1}{x}\right) \right\vert \le x^2$$
for $x \neq 0$. Consequently
$$\left\vert\frac{f(x)}{x} \right\vert \le \left\vert x \right\vert$$
which enables to prove that $f$ is differentiable at $0$ with $f^\prime(0)=0$.
For $x \neq 0$, you computed
$$f^\prime(x) = 2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$$
This answers questions a. and b. Remains questions c. and specifically continuity of $f^\prime$ at $0$.
Based on similar proofs than above you get:
$$\lim\limits_{x\to 0} 2\sin x \cos x \sin \frac{1}{x} =0$$ and you know that
$$\lim\limits_{x\to 0} \left(\frac{\sin x}{x}\right)^2 =1$$
You will conclude that $f^\prime$ is discontinuous at $0$ with an essential discontinuity at $0$ because $x \mapsto \cos\left(\frac{1}{x}\right)$ has itself an essential discontinuity at $0$.
A: Here you check whether this limit exists: $\lim_{x\to 0}2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$. The first term approaches zero, so we care about second (forget $-$ sign for some minuts):
$$\lim_{x\to 0}\frac{\cos(\frac{1}{x})\sin^2x}{x^2} \equiv \lim_{x\to 0}\cos(\frac{1}{x})$$
which surely does not exist. 
