Question

$\underset{n\to \infty }{\lim }\cos \left(\pi \sqrt{n^2+n}\right),n\in Z$ is equal to

• A. 0
• B. 1
• C. -1
• D. None of these

Answer 1

Answer: (A)

$\underset{n\to \infty }{\lim }\cos \left[\pi \sqrt{n^2+n}\right]$
$=\underset{n\to \infty }{\lim }\cos \left[n\pi {\left(1+\frac{1}{n}\right)}^{1/2}\right]$
$=\underset{n\to \infty }{\lim }\cos \left[n\pi \left(1+\frac{1}{2n}-\frac{1}{8n^2}+\dots \right)\right]$
$=\underset{n\to \infty }{\lim }\cos \left(n\pi +\frac{\pi }{2}-\frac{\pi }{8n}+\dots \right)$
$=-\underset{n\to \infty }{\lim }\sin \left(n\pi -\frac{\pi }{8n}+\dots \right)$
$=-\underset{n\to \infty }{\lim }(-1{)}^{n-1}\sin \left(\frac{\pi }{8n}-\dots \right)$
$(\because \frac{\pi }{8n}-\dots \to 0$ as $n\to \infty )$
$=0$