Question

Find the limit of $\frac{3x^3-2x^2-4}{5x^3-4x+8}$ when $\left(1\right)$ x is infinite$;$ $\left(2\right)$ x is zero.

Answer 1

To find the limit of

$\frac{3x^3-2x^2-4}{5x^3-4x+8}$

1) When $x$ is infinite

i.e. $\underset{x\to \infty }{lim}\frac{3x^3-2x^2-4}{5x^3-4x+8}$

$=\underset{x\to \infty }{lim}\frac{3-\frac{2}{x}-\frac{4}{x^3}}{5-\frac{4}{x^2}+\frac{8}{x^3}}$

$=\frac{3-0-0}{5-0-0}$

$=\frac{3}{5}$

2) When $x=0$

i.e. $\underset{x\to 0}{lim}\frac{3x^3-2x^2-4}{5x^3-4x+8}$

$=\frac{3(0{)}^3-2(0{)}^2-4}{5(0{)}^3-4(0)+8}$

$=\frac{-4}{8}$

$=\frac{-1}{2}$