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.

Solution 1:

HINT:

$$\frac{f(x^2)-f(0)}{x}=\left(\frac{f(x^2)-f(0)}{x^2}\right)x$$

Solution 2:

Recall the chain rule, $$ \left(f(g(x))\right)'=f'(g(x))g'(x) $$ here $g(x)=x^2$, and $$ \lim_{x\to 0}\frac{f(g(x))-f(g(0))}{x}=f'(g(0))(g'(0))=f'(0)\cdot 0=0 $$

Solution 3:

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*}