Question

$\underset{x\to \infty }{lim}(\sin \sqrt{x+1}-\sin \sqrt{x})$ is equal to -

• A. $1$
• B. $-1$
• C. $0$
• D. $-2$

Answer 1

Answer: (C)

$0$

${lim}_{x\to \infty }(\sin \sqrt{x+1}-\sin \sqrt{x})$

$={lim}_{x\to \infty }\left(2\sin \frac{\sqrt{x+1}-\sqrt{x}}{2}\cos \frac{\sqrt{x+1}+\sqrt{x}}{2}\right)$

$=2{lim}_{x\to \infty }\left(\sin \frac{\sqrt{x+1}-\sqrt{x}}{2}\cos \frac{\sqrt{x+1}+\sqrt{x}}{2}\right)$

Consider ${lim}_{x\to \infty }\sin \frac{\sqrt{x+1}-\sqrt{x}}{2}$

$={lim}_{x\to \infty }\frac{\sin \frac{\sqrt{x+1}-\sqrt{x}}{2}}{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\times \frac{\sqrt{x+1}-\sqrt{x}}{2}$

$=\frac{1}{2}{lim}_{x\to \infty }(\sqrt{x+1}-\sqrt{x})$

$=\frac{1}{2}{lim}_{x\to \infty }\frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{(\sqrt{x+1}+\sqrt{x})}$

$=\frac{1}{2}{lim}_{x\to \infty }\frac{x+1-x}{(\sqrt{x+1}+\sqrt{x})}$

$=\frac{1}{2}{lim}_{x\to \infty }\frac{1}{(\sqrt{x+1}+\sqrt{x})}$

$=\frac{1}{2}{lim}_{x\to \infty }\frac{1}{(\sqrt{x}\sqrt{1+\frac{1}{x}}+\sqrt{x})}$

$=\frac{1}{2}{lim}_{x\to \infty }\frac{1}{\sqrt{x}(\sqrt{1+\frac{1}{x}}+1)}$

$=\frac{1}{2}\times 0=0$

Now ,$2\cos \frac{\sqrt{x+1}+\sqrt{x}}{2}$ lies between finite values $-2$ and $2$ for all $x$ is finite.

Since the limits are finite, they can be multiplied and hence the limit is $0$