A) \[\infty \]
B) \[\frac{1}{2}\]
C) 2
D) 0
Correct Answer: B
Solution :
We have, \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1+2+3+.....+n}{{{n}^{2}}+100}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n(n+1)}{2({{n}^{2}}+100)}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{n}^{2}}\left( 1+\frac{1}{n} \right)}{2{{n}^{2}}\left( 1+\frac{100}{{{n}^{2}}} \right)}=\frac{1}{2}\].
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