The value of k which makes
Question:

The value of k which makes

$f(x)=\left\{\begin{array}{cc}\sin \frac{1}{x}, & x \neq 0 \\ k, & x=0\end{array}\right.$ continuous at $x=0$, is

(a) 8
(b) 1
(c) −1
(d) none of these

Solution:

(d) none of these

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0}\left(\sin \frac{1}{x}\right)=k$

But $\lim _{x \rightarrow 0}\left(\sin \frac{1}{x}\right)$ does not exist. Thus, there does not exist any $k$ that makes $f(x)$ a continuous function.

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