Question

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

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

Answer 1

Answer: (B)

$\frac{1}{2}$

Answer 2

Consider $\displaystyle\lim_{n \to\infty} \frac{ 1+2+3+ ...n}{n^{2} +100}$
$=\displaystyle\lim_{n\to\infty} \frac{n\left(n+1\right)}{\left(n^{2}+100\right)}$
(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 $D^r$.
$=\displaystyle\lim_{n\to\infty} \frac{n^{2} \left(1+ \frac{1}{n}\right)}{2n^{2}\left(1+ \frac{100}{n^{2}}\right) } = \frac{1}{2}$