Question

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

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

Answer 1

Answer: (A)

$\displaystyle\lim _{n \rightarrow \infty} \cos \left[\pi \sqrt{n^{2}+n}\right]$
$=\displaystyle\lim _{n \rightarrow \infty} \cos \left[n \pi\left(1+\frac{1}{n}\right)^{1 / 2}\right]$
$=\displaystyle\lim _{n \rightarrow \infty} \cos \left[n \pi\left(1+\frac{1}{2 n}-\frac{1}{8 n^{2}}+\ldots\right)\right]$
$=\displaystyle\lim _{n \rightarrow \infty} \cos \left(n \pi+\frac{\pi}{2}-\frac{\pi}{8 n}+\ldots\right)$
$=-\displaystyle\lim _{n \rightarrow \infty} \sin \left(n \pi-\frac{\pi}{8 n}+\ldots\right)$
$=-\displaystyle\lim _{n \rightarrow \infty}(-1)^{n-1} \sin \left(\frac{\pi}{8 n}-\ldots\right)$
$\left(\because \frac{\pi}{8 n}-\ldots \rightarrow 0\right.$ as $\left.n \rightarrow \infty\right)$
$=0$