Question

$\underset{x\to 0}{lim}\frac{\sqrt{1+{\tan }^{-1}3x}-\sqrt{1-{\sin }^{-1}3x}}{\sqrt{1-{\sin }^{-1}2x}-\sqrt{1+{\tan }^{-1}}2x}=$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (B)

$0$

We have,

${lim}_{x\to 0}\frac{\sqrt{1+{\tan }^{-1}3x}-\sqrt{1-{\sin }^{-1}3x}}{\sqrt{1-{\sin }^{-1}2x}-\sqrt{1+{\tan }^{-1}2x}}$

$\Rightarrow \frac{\sqrt{1+{\tan }^{-1}3\times 0}-\sqrt{1-{\sin }^{-1}3\times 0}}{\sqrt{1-{\sin }^{-1}2\times 0}-\sqrt{1+{\tan }^{-1}2\times 0}}$

$\Rightarrow \frac{\sqrt{1+{\tan }^{-1}0}-\sqrt{1-{\sin }^{-1}0}}{\sqrt{1-{\sin }^{-1}0}-\sqrt{1+{\tan }^{-1}0}}$

$\Rightarrow \frac{\sqrt{1+0}-\sqrt{1-0}}{\sqrt{1-0}-\sqrt{1+0}}$

$\Rightarrow 0$

Hence, this is the answer.