Functions Question 322
Question: If $ x_{n}=\frac{1-2+3-4+5-6+…..-2n}{\sqrt{n^{2}+1}+\sqrt{4n^{2}-1}}, $ then $ \underset{n\to \infty }{\mathop{\lim }},x_{n} $ is equal to
[AMU 2000]
Options:
A) $ \frac{1}{3} $
B) $ -\frac{2}{3} $
C) $ \frac{2}{3} $
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{n\to \infty }{\mathop{\lim }},\frac{1-2+3-4+5-6+…..-2n}{\sqrt{n^{2}+1}+\sqrt{4n^{2}-1}} $ $ f(x)=y $ $ =\frac{-2}{1+2}=\frac{-2}{3} $ .
BETA
