Question

If function f(x) = \begin{cases} x\, \sin\left(\frac{1}{x} \right) ; & \text{x \ne 0} \\\[2ex] a \,;& \text{x = 0} \end{cases} is continuous at $x = 0$, then value of $a$ is

• A. $1$
• B. $-1$
• C. $0$
• D. None of the options

Answer 1

Answer: (C)

$0$

Answer 2

We have, $f(0)=a$
$\therefore \displaystyle\lim _{x \rightarrow 0^{+}} f(x)=\displaystyle\lim _{h \rightarrow 0} f(0+h)=\displaystyle\lim _{h \rightarrow 0} h \sin \frac{1}{h}=0$
and $\displaystyle\lim _{x \rightarrow 0^{-}} f(x)=\displaystyle\lim _{h \rightarrow 0} f(0-h)$
$=\lim _{h \rightarrow 0}(-h) \sin \frac{1}{-h}=0$
Since, $f(x)$ is continuous at $x=0$, we must have
$f(0)=\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=\displaystyle\lim _{x \rightarrow 0^{-}} f(x) \Rightarrow a=0$