Question

Let $f:R\to R$ be a function such that $|f\left(x\right)|\le x^2$, for all $xϵR$. Then, 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

We have,
$-|f\left(x\right)|\le f\left(x\right)\le |f\left(x\right)|$

Let us consider the continuity of the function at $x=0$.

We have,

$\underset{x\to 0}{lim}-|f\left(x\right)|=0$ and

$\underset{x\to 0}{lim}|f\left(x\right)|=0$.

Hence, by applying the sandwich theorem,

$\underset{x\to 0}{lim}f\left(x\right)=0$

Hence, the function is continuous at $x=0$.
Now let us consider the differentiability at $x=0$.

We have,

$f^{\prime }\left(x\right)=\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x}$

$f^{\prime }\left(x\right)=\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}$

We have,

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

As $x\to 0$, the first and the third term tends to 0 .

Hence,
$f^{\prime }\left(x\right)=0$.
Hence, the function is differentiable at $x=0$