Question

Find the limits of the following expression $\frac{(3+2x^3)(x-5)}{(4x^3-9)(1+x)}$, $\left(1\right)$ when $x=\infty$, $\left(2\right)$ when $x=0$.

Answer 1

To find the limits of the following expression $\frac{(3+2x^3)(x-5)}{(4x^3-9)(1+x)}$

1) When $x=\infty$

i.e. $\underset{x\to \infty }{lim}\frac{(3+2x^3)(x-5)}{(4x^3-9)(1+x)}$

$=\underset{x\to \infty }{lim}\frac{\frac{(3+2x^3)(x-5)}{x^4}}{\frac{(4x^3-9)(1+x)}{x^4}}$

$=\underset{x\to \infty }{lim}\frac{(2+\frac{3}{x^3})(1-\frac{5}{x})}{(4-\frac{9}{x^3})(1+\frac{1}{x})}$

$=\frac{2\times 1}{4\times 1}$

$=\frac{1}{2}$

2) When $x=0$

i.e. $\underset{x\to 0}{lim}\frac{(3+2x^3)(x-5)}{(4x^3-9)(1+x)}$

$=\frac{(3+2(0{)}^3\left)\right(0-5)}{\left(4\right(0{)}^3-9\left)\right(1+0)}$

$=\frac{-15}{-9}$

$=\frac{5}{3}$