Question

Let $\left[x\right]$ denote the greatest integer less than or equal to $x$. Then the value of $\alpha$ for which the function $f\left(x\right)=\{\begin{array}{cc}\frac{\sin [-x^2]}{[-x^2]},& x\ne 0\\ \alpha ,& x=0\end{array}$

is continuous at $x=0$ is

• A. $\alpha =0$
• B. $\alpha =\sin (-1)$
• C. $\alpha =\sin \left(1\right)$
• D. $\alpha =1$

Answer 1

The correct option is C $\alpha =\sin \left(1\right)$

For $f\left(x\right)$ to be continuous at $x=0$,
$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$

LHL at $x=0$
$\underset{h\to 0}{lim}\frac{\sin [-(0-h{)}^2]}{[-(0-h{)}^2]}=\underset{h\to 0}{lim}\frac{\sin [-h^2]}{[-h^2]}=\frac{\sin (-1)}{-1}=\sin \left(1\right)$

RHL at $x=0$
$\underset{h\to 0}{lim}\frac{\sin [-(0+h{)}^2]}{[-(0+h{)}^2]}=\underset{h\to 0}{lim}\frac{\sin [-h^2]}{[-h^2]}=\frac{\sin (-1)}{-1}=\sin \left(1\right)$

$\therefore \underset{x\to 0}{lim}\frac{\sin [-x^2]}{[-x^2]}=\sin \left(1\right)$
$\therefore f\left(0\right)=\alpha =\sin \left(1\right)$