Question

$\underset{x\to 0}{lim}\frac{x+2\sin x}{\sqrt{x^2+2\sin x+1}-\sqrt{{\sin }^{2}x-x+1}}$ is :

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is B $2$

$\underset{x\to 0}{lim}\frac{x+2\sin x}{\sqrt{x^2+2\sin x+1}-\sqrt{{\sin }^{2}x-x+1}}=\underset{x\to 0}{lim}\frac{(x+2\sin x)(\sqrt{x^2+2\sin x+1}+\sqrt{{\sin }^{2}x-x+1})}{x^2+2\sin x+1-{\sin }^{2}x+x-1}=\underset{x\to 0}{lim}\frac{(x+2\sin x)}{x^2+2\sin x-{\sin }^{2}x+x}\times 2$
As it is $\frac{0}{0}$ form
Therefore, on applying L'Hospitals' Rule
$\underset{x\to 0}{lim}\frac{2(1+2\cos x)}{2x+2\cos x-2\sin x\cos x+1}=2$