Functions Question 318
Question: $ \underset{n\to \infty }{\mathop{\lim }},\sin [\pi \sqrt{n^{2}+1}]= $
Options:
A) $ \infty $
0
C) Does not exist
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Given limit $ =\underset{n\to \infty }{\mathop{\lim }},\sin { n\pi {{( 1+\frac{1}{n^{2}} )}^{1/2}} } $ $ =\underset{n\to \infty }{\mathop{\lim }},\sin ,{ n\pi ( 1+\frac{1}{2n^{2}}-\frac{1}{8n^{4}}+… ) } $
$ =\underset{x\to 0}{\mathop{\lim }},\frac{\frac{1}{3},{{(27-2x)}^{-2/3}}(-2)}{-\frac{3}{5},{{(243+5x)}^{-4/5}}(5)}=2. $
$ =\underset{n\to \infty }{\mathop{\lim }},{{(-1)}^{n}},\sin \pi ,( \frac{1}{2n}-\frac{1}{8n^{3}}+…. )=0. $
BETA
