Question

Evaluate the limit:
$\underset{x\to \infty }{lim}\frac{(3x-1)(4x-2)}{(x+8)(x-1)}$

Answer 1

We have,
$\underset{x\to \infty }{lim}\frac{(3x-1)(4x-2)}{(x+8)(x-1)}$

Dividing numerator and denominator by $x^2$, we get

$=\underset{x\to \infty }{lim}\frac{\frac{(3x-1)}{x}\frac{(4x-2)}{x}}{\frac{(x-8)}{x}\frac{(x-1)}{x}}$

$=\underset{x\to \infty }{lim}\frac{(3-\frac{1}{x})(4-\frac{2}{x})}{(1+\frac{8}{x})(1-\frac{1}{x})}$

$=\frac{\underset{x\to \infty }{lim}(3-\frac{1}{x})\underset{x\to \infty }{lim}(4-\frac{2}{x})}{\underset{x\to \infty }{lim}(1+\frac{8}{x})\underset{x\to \infty }{lim}(1-\frac{1}{x})}$

When $x\to \infty$, then $\frac{1}{x}\to 0$

$=\frac{(3-0)(4-2(0\left)\right)}{(1+8(0\left)\right)(1-0)}=12$