Question

Calculate the following limits.
$\underset{n\to \infty }{lim}\frac{\frac{1}{2}+1+\frac{3}{2}+...\frac{n}{2}}{0.25n^2+n+3},nϵN.$

Answer 1

$\underset{n\to \infty }{lim}\frac{\frac{1}{2}+1+\frac{3}{2}+...+\frac{n}{2}}{0.25n^2+n+3}$

$=\underset{n\to \infty }{lim}\frac{\frac{1}{2}(1+2+3+...+n)}{0.25n^2+n+3}$

$=\underset{n\to \infty }{lim}\frac{\frac{n(n+1)}{2}}{0.5n^2+2n+6}$

$=\underset{n\to \infty }{lim}\frac{n(n+1)}{n^2+4n+12}$

$=\underset{n\to \infty }{lim}\frac{n^2+n}{n^2+4n+12}$

$=\underset{n\to \infty }{lim}\frac{1+\frac{n}{n^2}}{1+4\frac{n}{n^2}+\frac{12}{n^2}}$

$=\underset{n\to \infty }{lim}\frac{1+\frac{1}{n}}{1+4\frac{1}{n}+\frac{12}{n^2}}$

$=\frac{1+0}{1+0+0}$

$=1$