Prove that $\lim_{x\rightarrow 0}\frac{f(x^2)-f(0)}{x}=0$ if $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable at $x=0$ Let the function $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable at $x=0$. Prove that $\lim_{x\rightarrow 0}\frac{f(x^2)-f(0)}{x}=0$.
The result is pretty obvious to me but I am having a difficult time arguing it precise enough for a proof. What I have so far is of course that since $f$ is differentiable;
$$f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}$$
exists.
Any help would be greatly appreciated.
 A: HINT:
$$\frac{f(x^2)-f(0)}{x}=\left(\frac{f(x^2)-f(0)}{x^2}\right)x$$
A: Recall the chain rule, 
$$
\left(f(g(x))\right)'=f'(g(x))g'(x)
$$
here $g(x)=x^2$, and 
$$
\lim_{x\to 0}\frac{f(g(x))-f(g(0))}{x}=f'(g(0))(g'(0))=f'(0)\cdot 0=0
$$
A: Given $\epsilon>0$, there is a $\delta>0$ such that
\begin{align*}
\left|\dfrac{f(u)-f(0)}{u}-f'(0)\right|<\epsilon/2,~~~~0<|u|<\delta,
\end{align*}
for all $x$ with $0<|x|<\min\{\sqrt{\delta},1,\epsilon/2(1+|f'(0)|)\}$, then
\begin{align*}
\left|\dfrac{f(x^{2})-f(0)}{x}\right|&=\left|x\left(\dfrac{f(x^{2})-f(0)}{x^{2}}-f'(0)\right)+xf'(0)\right|\\
&\leq|x|\left|\dfrac{f(x^{2})-f(0)}{x^{2}}-f'(0)\right|+|x||f'(0)|\\
&\leq\epsilon.
\end{align*}
A: If $g(x) = f(x^2)$, then, by the chain rule, $\lim\limits_{x\to 0}\frac{g(x)-g(0)}{x} = \color{blue}{g'(0)} = 2 (\color{red}{0}) f'(0) = \color{red}{0}$.
A: To add a little bit of intuition here, consider trying to construct a counterexample.  To obtain a limit of 1, instead of 0, we could construct a function where $f(x^2)-f(0) = x$ for all x, and the equation becomes $x/x = 1$ everywhere. The function $f(x) = \sqrt x$ satisfies this for positive x.  Allowing negative numbers and ensuring continuity around zero is then satisfied with $f(x) = \begin{cases} \sqrt x  & \text{if } x\ge 0 \\
                      - \sqrt -x & \text{if } x\lt 0      %
        \end{cases}$
Indeed the limit as x approaches zero is $x / x = 1$.  However, it's not a disproof, because in constructing this function, we had to violate the requirements. It has a slope of infinity at $f(0)$ and therefore is not regarded as differentiable at this point, although it is smooth and continuous.  Intuitively, any function which tries to violate the stated limit equation must be discontinuous or with infinite slope at zero.
