_ x 0 x x+1-1-x is equal to - Vidyadip

MCQ

$\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}}$ is equal to

✓
$1$
B
$0$
C
$2$
D
$-1$

✓

Answer

Correct option: A.

$1$

Given, $\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}}$
$=\lim _{x \rightarrow 0} \frac{\sin x[\sqrt{x+1}+\sqrt{1-x}]}{(\sqrt{x+1}-\sqrt{1-x}) \cdot(\sqrt{x+1}+\sqrt{1-x})}$
$=\lim _{x \rightarrow 0} \frac{\sin x[\sqrt{x+1}+\sqrt{1-x}]}{x+1-1+x}$
$=\lim _{x \rightarrow 0} \frac{\sin x[\sqrt{x+1}+\sqrt{1-x}]}{2 x}$
$=\frac{1}{2} \cdot \lim _{x \rightarrow 0} \frac{\sin x}{x}[\sqrt{x+1}+\sqrt{1-x}]$
Taking limits, we get
$=\frac{1}{2} \times 1 \times[\sqrt{0+1}+\sqrt{1-0}]$
$=\frac{1}{2} \times 1 \times 2$
$=1$

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