Question

$\underset{n\to \infty }{lim}\left(\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1}\right)$ is equal to.

• A. $0$
• B. $1$
• C. $1/5$
• D. $\infty$

Answer 1

The correct option is C $1/5$

$\underset{n\to \infty }{lim}\left(\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1}\right)$

$=\underset{n\to \infty }{lim}\left(\frac{2n^3(5n+1)+(1-5n^2)(2n^2+3)}{(2n^2+3)(5n+1)}\right)$

$=\underset{n\to \infty }{lim}\left(\frac{10n^4+2n^3+2n^2-10n^4+3-15n^2}{(2n^2+3)(5n+1)}\right)$

$=\underset{n\to \infty }{lim}\left(\frac{2n^3-13n^2+3}{10n^3+2n^2+15n+3}\right)$

$=\underset{n\to \infty }{lim}\left(\frac{2-\frac{13}{n}+\frac{3}{n^3}}{10+\frac{2}{n}+\frac{15}{n^2}+\frac{3}{n^3}}\right)$

$=\frac{2}{10}=\frac{1}{5}$

$\therefore \underset{n\to \infty }{lim}\left(\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1}\right)=\frac{1}{5}$