Functions Question 321
Question: The value of $ \underset{n,\to ,\infty }{\mathop{\lim }},\frac{1-n^{2}}{\sum n} $ will be
[UPSEAT 1999]
Options:
A) ? 2
B) ? 1
C) 2
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{n\to \infty }{\mathop{\lim }}\frac{1-n^{2}}{\Sigma n} $ $ =\underset{n\to \infty }{\mathop{\lim }},\frac{(1-n)(1+n)}{\frac{1}{2}n(n+1)} $ $ =\underset{n\to \infty }{\mathop{\lim }}\frac{2,(1-n)}{n} $ $ =\underset{n\to \infty }{\mathop{\lim }},2,( \frac{1}{n}-1 ) $ $ =2(0-1)=-2 $ .
BETA
