Question

2x^3-3x^2-9x/x^3-9x
lim x --3

Answer 1

$\lim_{x \to -3} \frac{2x^3-3x^2-9x}{x^3-9x} = \lim_{x \to -3} \frac{x(2x^2-3x-9)}{x(x^2-9)}=\lim_{x \to -3} \frac{(2x^2-3x-9)}{(x-3)(x+3)}\\\implies(2x^2-3x-9)\implies \\ ( \frac{3\pm \sqrt{81} }{4}=x'= \frac{3-9}{4}=-1,5|\ x''= \frac{3+9}{4}=3)\implies (x+ \frac{1}{2})(x-3))\\\\ \lim_{x \to -3} \frac{(x+\frac{1}{2})(x-3)}{(x-3)(x+3)}= \lim_{x \to -3} \frac{x+\frac{1}{2}}{x+3}= \frac{-3+\frac{1}{2}}{-3+3} =\\-\infty \ se\ \lim_{x \to -3^-} \\\ \ \infty \ se\ \lim_{x \to -3^+}$