Question

Let $f\left(x\right)=\sin \frac{1}{x},x\ne 0$. Then $f\left(x\right)$ can be continuous at $4x=0$

• A. If $f\left(0\right)=1$
• B. If $f\left(0\right)=0$
• C. If $f\left(0\right)=-1$
• D. For no definite value of $f\left(0\right)$

Answer 1

The correct option is D For no definite value of $f\left(0\right)$

For continuity at $x=0$

$\underset{h\to 0+}{lim}\sin \frac{1}{0+h}=\underset{h\to 0-}{lim}\sin \frac{1}{0-h}=f\left(0\right)$

But $\underset{h\to 0+}{lim}\sin \frac{1}{0+h}$ and $\underset{h\to 0-}{lim}\sin \frac{1}{0-h}$ are any number oscillating between $1$ and $-1$ Which are not equal to a defined number.

$\therefore f\left(0\right)$ cannot be a definite number.