Question

For $x\in R$, let $f\left(x\right)=\{\begin{array}{cc}{x}^3\sin \left(\frac{1}{x}\right),& x\ne 0\\ 0,& x=0\end{array}.$ Then which of the following options is (are) TRUE?

• A. $f$ is differentiable at $x=0$
• B. $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^2}=0$
• C. $\frac{f\left(x\right)}{x^3}$ is continuous at $x=0$
• D. $\underset{x\to \infty }{lim}\frac{f\left(x\right)}{x^2}=1$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $f$ is differentiable at $x=0$
B $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^2}=0$
D $\underset{x\to \infty }{lim}\frac{f\left(x\right)}{x^2}=1$

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}{x}^3\sin \left(\frac{1}{x}\right)$
$=0\left[\text{By sandwich theorem of limits}\right]$
$=f\left(0\right)$
Hence, $f$ is continuous at $x=0$.

Checking differentiability at $x=0:$
$f^{\prime }\left(0\right)=\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}$
$=\underset{x\to 0}{lim}\frac{x^3\sin \left(\frac{1}{x}\right)}{x}$
$=\underset{x\to 0}{lim}{x}^2\sin \left(\frac{1}{x}\right)$
$=0$
Hence, $f$ is differentiable at $x=0$.


$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^2}$
$=\underset{x\to 0}{lim}x\sin \left(\frac{1}{x}\right)$
$=0\left[\text{By sandwich theorem of limits}\right]$

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}$
$=\underset{x\to 0}{lim}\sin \left(\frac{1}{x}\right)$ which does not exist.
Hence, $\frac{f\left(x\right)}{x^3}$ is not continuous at $x=0$.

$\underset{x\to \infty }{lim}\frac{f\left(x\right)}{x^2}$
$=\underset{x\to \infty }{lim}x\sin \left(\frac{1}{x}\right)$
$=\underset{h\to 0}{lim}\frac{\sin h}{h}[\text{Putting}x=\frac{1}{h}]$
$=1$