Prove or disprove: if $|f(x)|\leq x^2$ then $f$ is differentiable at $0$ I'm trying to prove/disprove the following statement:

If $|f(x)|\leq x^2$ for all $x$, then $f$ is differentiable at $x=0$.

My initial attempt: $|f(0)|\leq0\Rightarrow f(0)=0$. Then, because $-x^2\leq f(x)\leq x^2$ we get that $f'(0)=\lim\limits_{x\to0} \frac{f(x)-0}{x-0}=0$ (using Squeeze rule). But it appears to be wrong and I can't find a counter example. Any suggestions?
 A: $$\left|\frac{f(x)-f(0)}{x}\right|\le\left|\frac{x^2-0}{x}\right|=|x|\to 0$$
as $x\to 0$  The limit of the left-hand side is $f'(0)=0$.
A: Dr. MV's solution is correct; Here, I am only willing to comment a few occasions where we could/couldn't prove the result.
Remark 1: If there existed a $\lambda > 1$ such that $|f(x)|\le |x|^{\lambda}$, the result would follow by the same reasoning. This has to do with the Hölder Cotinuity exponent of the function at $x=0$. 

Def.: We call $f : \Bbb{R} \rightarrow \Bbb{R} $Hölder continuous of exponent $\lambda > 0$ at the point $x_0$ if we can assure that the inequality $|f(y)-f(x_0)|\le C |y-x_0|^{\lambda}$ for all $y$. ($C$ is a fixed constant)
We then call a function *Hölder continuous of exponent*$\lambda$ if the above condition holds at every $x_0$.

It can be (easily) showed that the only Hölder continuous functions of exponent $\lambda > 1$ are the constant functions. For $\lambda = 1$, we recover the concept of Lipschitz functions - that are far from being 'trivial' in the former sense, but which are 'essentially' differentiable functions -, and for $0<\lambda <1$ we have a wide class of non-differentiable functions - for example, a wide class of irregular continuous functions, such as the Brownian Motion's paths, are Hölder Continuous of some exponent. 
Remark 2: The previous remark (maybe) suggests us that the whole thing must fail even for $\lambda = 1$. For example, $f(x)=|x|$ is not even differentiable at $x=0$. 
For example, the Brownian Motion's paths are locally Hölder Coninuous of exponent $\forall \; \alpha <1/2$, but they are nowhere differentiable almost surely. This provides some more examples of how can a function be relatively well-behaved but not differentiable. 
