Question

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

Answer 1

To find the limits of the following expression

$\frac{1-x^2}{2x^3-1}÷\frac{1-x}{2x^2}$

1) When $x=\infty$

$\frac{1-x^2}{2x^3-1}÷\frac{1-x}{2x^2}$

$=\frac{1-x^2}{2x^3-1}\times \frac{2x^2}{1-x}$

$=\frac{(1-x)(1+x)}{2x^3-1}\times \frac{2x^2}{1-x}$

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

$=\frac{2x^2+2x^3}{2x^3-1}$

Now,

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

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

$=\frac{2}{2}$

$=1$

2) When $x=0$

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

$=\frac{2\times (0{)}^2+2\times (0{)}^3}{2\times (0{)}^3-1}$

$=\frac{0}{-1}$

$=0$