Question

Solve : $\underset{x\to \frac{\pi }{2}}{lim}\frac{\sqrt{3+sinx}-2}{{cos}^{2}x}$

Answer 1

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

$=\frac{0}{0}$ form

Rationalizing the numerator.

$\Rightarrow \underset{x\to \frac{\pi }{2}}{lim}\frac{\sqrt{3+\sin x}-2}{{\cos }^{2}x}\times \frac{\sqrt{3+\sin x}+2}{\sqrt{3+\sin x}+2}$

$=\underset{x\to \frac{\pi }{2}}{lim}\frac{(3+\sin x)-4}{{\cos }^{2}x(\sqrt{3+\sin x}+2)}$

$=\underset{x\to \frac{\pi }{2}}{lim}\frac{\sin x-1}{(1-{\sin }^{2}x)(\sqrt{3+\sin x}+2)}$ (using ${\cos }^{2}x=1-{\sin }^{2}x$)

$=\underset{x\to \frac{\pi }{2}}{lim}\frac{-(1-\sin x)}{(1-\sin x)(1+\sin x)(\sqrt{3+\sin x}+2)}$

$=\underset{x\to \frac{\pi }{2}}{lim}\frac{-1}{(1+\sin x)(\sqrt{3+\sin x}+2)}$

$=\frac{-1}{(1+1)(\sqrt{3+1}+2)}$

$=\frac{-1}{2\left(4\right)}$

$=-\frac{1}{8}$.