Question

Let $f\left(x\right)=\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}$. The value which should be assigned to $f$ at $x=0$ so that it is continuous everywhere is-

• A. $1$
• B. $2$
• C. $-2$
• D. $1/2$

Answer 1

The correct option is A $1$

$f\left(0\right)=\underset{x\to 0}{lim}\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}$
$=\underset{x\to 0}{lim}\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}.\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}$
$=\underset{x\to 0}{lim}\frac{2\sin x}{x}.\frac{1}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}=\frac{2}{2}=1$