# If $\left|f(x)\right|\leq x^2\;\forall x \in \mathbb{R}$. Then Cont. and Diff. at $x=0$

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function such that $\left|f(x)\right|\leq x^2\;\forall x \in \mathbb{R}$. Then how can we prove that

function $f(x)$ is Continuous and Differentiable at $x=0.$

$\bf{My\; Try::}$ Given $\left|f(x)\right|\leq x^2\; \forall x \in \mathbb{R}.$ So we put $x=0$

We get $\left|f(0)\right|\leq 0^2\Rightarrow f(0) = 0$.

Now We will calculate for Differentiability of function at $x=0.$ bcz If function is Differentiable

at $x=0$. Then it must be Cont. at $x=0$

So Given $\displaystyle \left|f(x)\right|\leq x^2\Rightarrow \left|\frac{f(x)-f(0)}{x-0}\right|\leq |x|$

Now How Can I solve after that, Help me

Thanks

Just calculate $f'(0)$ by the definition of $f'$, i.e.

$$f'(0) = \lim_{h\to 0}\frac{f(h)-f(0)}{h-0}$$

This limit exists because, as you showed,

$$\left|\frac{f(x)-f(0)}{x-0}\right|\leq |x|.$$

You know that

$$0\le\left|\frac{f(x)-f(0)}{x-0}\right|\leq |x|$$

Now just use the squeeze theorem for $x\to 0$ to get that $f'(0)$ exists and is $0$.