Question:
Evaluate the following limits:
$\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\cos x}}{x}$
Solution:
$=\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}$
$=\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x} \times \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}$
$=\lim _{x \rightarrow 0} \frac{1+\sin x-(1-\sin x)}{x(\sqrt{1+\sin x}+\sqrt{1-\sin x})}$
$=\lim _{x \rightarrow 0} \frac{2 \times \sin x}{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}$
$=2 \times \lim _{x \rightarrow 0} \frac{\sin x}{x} \times \lim _{x \rightarrow 0} \frac{1}{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}$
$=2 \times 1 \times \frac{1}{2}$
$=1$
$\therefore \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}=1$