Question

Find the value of m for which lim x->infinity

(x+1)(2x-3)(2-3x)/(4-x+mx³) exists.

Answer 1

$\lim_{x\to\infty} \frac{(x+1)(2x-3)(2-3x)}{(4-x+mx^3)}\\ =\lim_{x\to\infty} \frac{-6x^3+7x^2+7x-6}{(4-x+mx^3)}\\$

Divide numerator and denominator by $x^3$

$=\lim_{x\to\infty} \frac{-6+\frac{7}{x}+\frac{7}{x^2}-\frac{6}{x^3}}{(\frac{4}{x^3}-\frac{1}{x^2}+m)}\\ =\frac{-6}{m}$

For limit to be exist, $m\neq0$