Question

$\underset{n\to \infty }{\lim }\frac{1+2+3+...+n}{n^2+100}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$$\begin{gathered} \begin{array}{c}\lim \\ n\to \infty \end{array}\frac{1+2+..........+n}{n^2+100} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ n\to \infty \end{array}\frac{{\displaystyle \frac{n\left(n+1\right)}{2}}}{n^2+100}\left(\because \mathrm{Sum}\mathrm{of}\mathrm{first}n\mathrm{natural}\mathrm{number}\mathrm{is}\frac{n\left(n+1\right)}{2}\right) \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ n\to \infty \end{array}\frac{1}{2}\left[\frac{n^2+n}{n^2+100}\right] \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ n\to \infty \end{array}\frac{1}{2}\left[\frac{1+{\displaystyle \frac{1}{n}}}{1+{\displaystyle \frac{100}{n^2}}}\right]\left(\mathrm{By}\mathrm{dividing}\mathrm{by}{n}^2\mathrm{both}\mathrm{numerator}\mathrm{of}\mathrm{denominator}\right) \\ \left(\mathrm{Since}\begin{array}{c}\lim \\ n\to \infty \end{array}\frac{1}{n}=0\right) \end{gathered}$$
$\begin{array}{rcl}\mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ n\to \infty \end{array}\frac{1}{2}\left(\frac{n\left(n+1\right)}{n^2+100}\right)& =& \frac{1}{2}\frac{\left(1+\begin{array}{c}\lim \\ n\to \infty \end{array}{\displaystyle \frac{1}{n}}\right)}{1+\begin{array}{c}\lim \\ n\to \infty \end{array}{\displaystyle \frac{100}{n^2}}}\\ & =& \frac{1}{2}\frac{\left(1+0\right)}{1+0}\end{array}$
$\mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ n\to \infty \end{array}\frac{1+2+.....+n}{n^2+100}=\frac{1}{2}$