Question

Find the value of the following limit:
$\underset{x\to \infty }{lim}\frac{2x^2+7x-2}{6x^3-4x+3}$

Answer 1

$\underset{x\to \infty }{lim}\frac{2x^2+7x-2}{6x^3-4x+3}$

$=\underset{x\to \infty }{lim}\frac{x^2(2+\frac{7}{x}-\frac{2}{x^2})}{x^3(6-\frac{4}{x^2}+\frac{3}{x^3})}$

$=\underset{x\to \infty }{lim}\frac{1}{x}\left(\frac{2+\frac{7}{x}-\frac{2}{x^2}}{6-\frac{4}{x^2}+\frac{3}{x^3}}\right)$

$=\underset{x\to \infty }{lim}\frac{\frac{1}{\infty }(2+\frac{7}{\infty }-\frac{2}{\infty })}{(6-\frac{4}{\infty }+\frac{3}{\infty })}$

$=\frac{0\times (2+0-0)}{(6-0+0)}=0$.