Functions Question 309
Question: $
[.] $ is equal to [IIT 1984; DCE 2000; Pb. CET 2000]
Options:
A) 0
B) $ -\frac{1}{2} $
C) $ \log ( \frac{2}{3} ) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{n\to \infty }{\mathop{\lim }}[ \frac{1}{1-n^{2}}+\frac{2}{1-n^{2}}+…..+\frac{n}{1-n^{2}} ] $ $ =\underset{n\to \infty }{\mathop{\lim }}\frac{\Sigma n}{1-n^{2}}=\frac{1}{2}\underset{n\to \infty }{\mathop{\lim }}\frac{n^{2}+n}{1-n^{2}}=-\frac{1}{2} $ .
BETA
