Question

Evaluate:
$\rm \displaystyle \lim_{x \to \infty}\ \frac{(3x+4)(4x-6)(5x+2)}{4x^{3}+2x^{2}-1}$

Answer 1

Solution :

$\rm \displaystyle \lim_{x \to \infty}\ \frac{(3x+4)(4x-6)(5x+2)}{4x^{3}+2x^{2}-1}$

Divide numerator and denominator by

x³ , we get

= $\rm \displaystyle \lim_{x \to \infty}\ \frac{x^{3}[(3+\frac{4}{x})(4-\frac{6}{x})(5+\frac{2}{x})]}{x^{3}[(4+\frac{2}{x}-\frac{1}{x^3})]}$

= [ ( 3+0)(4-0)(5+0)]/[ 4+0-0 ]

= ( 3 × 4 × 5 )/4

= 3 × 5

= 15

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