Question

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

Answer 1

To find the limits of the following expression

$\frac{(x-3)(2-5x)(3x+1)}{(2x-1{)}^3}$

1) When $x=\infty$

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

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

$=\frac{1\times (-5)\times \left(3\right)}{2^3}$

$=\frac{-15}{8}$

2) When $x=0$

$\underset{x\to 0}{lim}\frac{(x-3)(2-5x)(3x+1)}{(2x-1{)}^3}$

$=\frac{(0-3)(2-5\times 0)(3\times 0+1)}{(2\times 0-1{)}^3}$

$=\frac{-6}{-1}$

$=6$