Functions Question 326
Question: The value of $ \underset{n\to \infty }{\mathop{\lim }},\frac{1+2+3+….n}{n^{2}+100} $ is equal
[Pb. CET 2002]
Options:
A) $ \infty $
B) $ \frac{1}{2} $
C) 2
D) 0
Show Answer
Answer:
Correct Answer: B
Solution:
We have, $ \underset{n\to \infty }{\mathop{\lim }},\frac{1+2+3+…..+n}{n^{2}+100} $ $ =\underset{n\to \infty }{\mathop{\lim }},\frac{n(n+1)}{2(n^{2}+100)}=\underset{n\to \infty }{\mathop{\lim }},\frac{n^{2}( 1+\frac{1}{n} )}{2n^{2}( 1+\frac{100}{n^{2}} )}=\frac{1}{2} $ .
BETA
