Prove that $f(x)$ is differentiable at any point Given this condition:
$x^2 \sin(x^2) \le x^3f(x)\le \sin(x^4)$ at any open interval that goes through $0$
I need to prove that $f(x)$ is differentiable at $x=0$, but I couldn't come up how, 
it looks like the condition above could help me somehow use the squeeze theorem, but I'm not sure.
Any help
 A: You have that $$\frac{\sin(x^2)}{x}\leq f(x)\leq\frac{\sin(x^4)}{x^3}$$
for all $x\neq0$. Taking limits when $x\to 0$ and assuming $f$ is continuous at $x=0$, we get $$0\leq f(0)\leq0$$
Therefore $f(0)=0$.
Nor we compute the direvative from the definition. We have
$$1=\lim_{x\to0^+}\frac{\sin(x^2)}{x^2}\leq \lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=f'_+(0)\leq\lim_{x\to0^+}\frac{\sin(x^4)}{x^4}=1$$
and $$1=\lim_{x\to0^-}\frac{\sin(x^2)}{x^2}\geq \lim_{x\to0^-}\frac{f(x)-f(0)}{x-0}=f'_-(0)\geq\lim_{x\to0^-}\frac{\sin(x^4)}{x^4}=1$$
Therefore $f_+'(0)=f_-'(0)=f'(0)=1$.

The continuity assumption cannot be avoided. Otherwise you can redefine $f(0)$ while keeping the validity of the inequality, but losing the differentiability at $x=0$. 

A: Divide all sides by $x^4$ and use the fact that $\displaystyle \lim_{u\to 0} \frac{\sin u}{u} =1$ to get that, by the squeeze theorem, $\displaystyle \lim_{x\to 0} \frac{f(x)}{x}=1$. The result follows if $f(0)=0$ by definition of the derivative. However, it doesn't follow if $f(0)\neq 0$. I don't think that we can deduce this value of $f(0)$ from the given data.
A: The given inequality implies f(0) = 0. Now, divide everything by x^4, provided x non-zero, then take x to 0. You will get that f is differentiable and f '(0) = 1.
