Prove that f is differentiable at x=0 Suppose $0 &lt; f(x) &lt; x^2, \ \forall x$. Prove that $f$ is differentiable at $x=0$ and $f'(x)=0$.

I don't even know how to start with the first one. For the second question, I'm guessing we can use squeeze theorem to show that the limit must also go to $0$ as the limit of the bounds goes to 0. 

Much appreciated!
I take it you mean $0 \leq f(x)$. It is easy to see that $f(0) = 0$. 

For $h\neq 0$, we have $0 \leq f(h) \leq h^2$. 

If $$h&gt;0 \Longrightarrow  0 \leq \frac{f(h)}{h} \leq h$$ $$h&lt;0 \Longrightarrow h \leq \frac{f(h)}{h} \leq 0$$

Observe that these are the difference quotients for the three functions at $x=0$. Now squeeze.
Once we have $ \ f: \mathbb{R} \to \mathbb{R} \ $ such that $ \ 0 \leq f(x) \leq x^2$, $\forall x \in \mathbb{R}$, is straightforward that $ \ f(0)=0$. Then, $\forall t \in \mathbb{R} \setminus \{ 0 \}$, we have that $$\displaystyle 0 \leq \frac{f(t)}{t} \leq \frac{t^2}{t} = t$$ By the Sandwich theorem, we conclude that $$\displaystyle f'(0) = \lim_{t \to 0} \frac{f(0+t) - f(0)}{t} = \lim_{t \to 0} \frac{f(t)}{t} = 0$$