Question

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

• A. 2
• B. -2
• C. 0
• D. None of these

Answer 1

Answer: (C)

0

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

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

Now for

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

$=\underset{x\to \infty }{lim}\frac{\sin \left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right)}{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\times \frac{\sqrt{x+1}-\sqrt{x}}{2}$ (sandwich theorem)

$=\underset{x\to \infty }{lim}\frac{\sqrt{x+1}-\sqrt{x}}{2}$

$=\frac{1}{2}\underset{x\to \infty }{lim}\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}\cdot \frac{\sqrt{x+1}+\sqrt{x}}{1}$

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

Now, $2\cos \left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)$ is always finite and lies between $[-2,2]$

$\therefore \underset{x\to \infty }{lim}2\cos \left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)\sin \left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right)=finite\times 0$

$=0$