Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

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

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

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

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

$=1\times \frac{1}{2}=\frac{1}{2}$