Question

Find the limits of the following expression $\frac{(3x^2-1{)}^2}{x^4+9}$, $\left(1\right)$ when $x=\infty ,$ $\left(2\right)$ when $x=0$.

Answer 1

To find the limits of the following expression

$\frac{(3x^2-1{)}^2}{x^4+9}$

1) When $x=\infty$

i.e. $\underset{x\to \infty }{lim}\frac{(3x^2-1{)}^2}{x^4+9}$

$=\underset{x\to \infty }{lim}\frac{9x^4+1-6x^2}{x^4+9}$

$=\underset{x\to \infty }{lim}\frac{9+\frac{1}{x^4}-\frac{6}{x^2}}{1+\frac{9}{x^4}}$

$=9$

2) When $x=0$

$\underset{x\to 0}{lim}\frac{(3x^2-1{)}^2}{x^4+9}$

$=\underset{x\to 0}{lim}\frac{9x^4+1-6x^2}{x^4+9}$

$=\frac{1}{9}$