Question

Find the limits of the following expression $\frac{(2x-3)(3-5x)}{7x^2-6x+4}$, $\left(1\right)$ when $x=\infty ,$ $\left(2\right)$ when $x=0$.

Answer 1

To find the limits of the following expression

$\frac{(2x-3)(3-5x)}{7x^2-6x+4}$

1) When $x=\infty$

$\underset{x\to \infty }{lim}\frac{(2x-3)(3-5x)}{7x^2-6x+4}$

$\underset{x\to \infty }{lim}\frac{\frac{(2x-3)(3-5x)}{x^2}}{\frac{7x^2-6x+4}{x^2}}$

$\underset{x\to \infty }{lim}\frac{(2-\frac{3}{x})(\frac{3}{x}-5)}{7-\frac{6}{x}+\frac{4}{x^2}}$

On Applying limit , we get

$=\frac{\left(2\right)(-5)}{7}$

$=\frac{-10}{7}$

2 ) When $x=0$

$\underset{x\to 0}{lim}\frac{(2x-3)(3-5x)}{7x^2-6x+4}$

$=\frac{(2\times 0-3)(3-5\times 0)}{7(0{)}^2-6(0)+4}$

$=\frac{(-3)\left(3\right)}{4}$

$=\frac{-9}{4}$