Question

Find the following limit:
$\underset{x\to \infty }{lim}\frac{x^2+3x-4}{1-3x^2}.$

Answer 1

$\underset{x\to \infty }{lim}\frac{x^2+3x-4}{1-5x^2}$

$=\underset{x\to \infty }{lim}\frac{x^2(1+\frac{3}{x}-\frac{4}{x^2})}{x^2(\frac{1}{x^2}-5)}$

$=\frac{1+\frac{3}{\infty }-\frac{4}{{\infty }^2}}{\frac{1}{{\infty }^2}-5}[\because \frac{1}{\infty }=0]$

$=\frac{1+0-0}{0-5}=\frac{-1}{5}$.