Find the value of m for which lim x->infinity
(x+1)(2x-3)(2-3x)/(4-x+mx³) exists.
"\\lim_{x\\to\\infty} \\frac{(x+1)(2x-3)(2-3x)}{(4-x+mx^3)}\\\\\n=\\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)}\\\\\n=\\frac{-6}{m}"
For limit to be exist, "m\\neq0"
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