Question

$\underset{x\to 0}{lim}\frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}}$ is

• A. $2$
• B. $0$
• C. $1$
• D. $-1$

Answer 1

Answer: (C)

$1$

$\underset{x\to 0}{lim}\frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}}$

First we are puting $x\to 0$ then we get Indeterminate form $\frac{0}{0}$.

Now, we do rationalization.

$\underset{x\to 0}{lim}\frac{\sin x\times (\sqrt{x+1}+\sqrt{1-x})}{(\sqrt{x+1}-\sqrt{1-x})\times (\sqrt{x+1}+\sqrt{1-x})}$

$\Rightarrow \underset{x\to 0}{lim}\frac{\sin x\times (\sqrt{x+1}+\sqrt{1-x})}{2x}$.

$\Rightarrow \underset{x\to 0}{lim}\frac{\sin x}{x}\times \underset{x\to 0}{lim}\frac{(\sqrt{x+1}+\sqrt{1-x})}{2}$.

$[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$\Rightarrow 1\times \underset{x\to 0}{lim}\frac{(\sqrt{x+1}+\sqrt{1-x})}{2}$

$1\times 1=1$