Question

The value of $\underset{n\to \infty }{\lim }\frac{1+2+3+...n}{n^2+100}$ is equal to :

• A. $\infty$
• B. $\frac{1}{2}$
• C. 2
• D. 0

Answer 1

Answer: (B)

Consider $\underset{n\to \infty }{\lim }\frac{1+2+3+...n}{n^2+100}$
$=\underset{n\to \infty }{\lim }\frac{n\left(n+1\right)}{\left(n^2+100\right)}$
(By using sum of $n$ natural number $1+2+3+....+n=\frac{n\left(n+1\right)}{2}$)
Take $n^2$ common from $N^r$ and $D^r$.
$=\underset{n\to \infty }{\lim }\frac{n^2\left(1+\frac{1}{n}\right)}{2n^2\left(1+\frac{100}{n^2}\right)}=\frac{1}{2}$