Question

The value of $\underset{n\to \infty }{lim}\frac{\frac{1}{2}+1+\frac{3}{2}+....\frac{n}{2}}{25n^2+n+3}$ is

• A. $0$
• B. $\frac{1}{100}$
• C. $\infty$
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{100}$

$\underset{n\to \infty }{lim}\frac{\frac{1}{2}+1+\frac{3}{2}+....\frac{n}{2}}{25n^2+n+3}$
$=\underset{n\to \infty }{lim}\frac{\frac{n}{2}(2.\frac{1}{2}+(n-1\left)\frac{1}{2}\right)}{25n^2+n+3}$
$=\underset{n\to \infty }{lim}\frac{\frac{n}{2}\left(\frac{n+1}{2}\right)}{25n^2+n+3}$
$=\frac{1}{4}\underset{n\to \infty }{lim}\frac{n(n+1)}{25n^2+n+3}$
$=\frac{1}{4}\underset{n\to \infty }{lim}\frac{(1+1/n)}{25+1/n+3/n^2}=\frac{1}{100}$