Question

Let $f:R\to R$ be a function such that $|f\left(x\right)|\le x^2$, for all $x\in R$. At $x=0,f$ is

• A. Continuous but not differentiable
  
• B. Continuous as well as differentiable
• C. Neither continuous nor differentiable
• D. Differentiable but not continuous

Answer 1

The correct option is B Continuous as well as differentiable

Since, $|f\left(x\right)|\le x^2\forall x\in R$ , we have at $x=0,|f\left(0\right)|\le 0$
$\therefore f\left(0\right)=0...\left(1\right)$

$\therefore f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$
$\Rightarrow f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}...\left(2\right)$

Now, $\left|\frac{f\left(h\right)}{h}\right|\le |h|(\because |f(x)|\le x^2)$

$\Rightarrow -|h|\le \frac{f\left(h\right)}{h}\le |h|$
Applying limit on the above equation
$-\underset{h\to 0}{lim}|h|\le \underset{h\to 0}{lim}\frac{f\left(h\right)}{h}\le \underset{h\to 0}{lim}|h|$

By using sandwich theorem,
$\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}=0...\left(3\right)$

Therefore, from eqn$\left(2\right)$ and $\left(3\right)$, we get $f^{\prime }\left(0\right)=0$ i.e., $f\left(x\right)$ is differentiable at $x=0$.