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