Question

The value of k which makes $f(x) = \left\{ \begin{aligned} & \begin{matrix} \sin\left( \frac{1}{x} \right), & x \neq 0 \end{matrix} \\ & \begin{matrix} k, & x = 0 \end{matrix} \end{aligned} \right.\$ continuous at x = 0 is

• A. 8
• B. 1
• C. –1
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

We have $\lim_{x \rightarrow 0}f(x) = \lim_{x \rightarrow 0}\sin\frac{1}{x}$= An oscillating number which oscillates between –1 and 1.

Hence, $\lim_{x \rightarrow 0}f(x)$ does not exist. Consequently $f(x)$ cannot be continuous at $x = 0$ for any value of k.