Answer
$\lim\limits_{x \to -\infty}(x^2-x^3)e^{2x} = 0$
Work Step by Step
$\lim\limits_{x \to -\infty}(x^2-x^3)e^{2x}$
$=\lim\limits_{x \to -\infty}\frac{x^2-x^3}{e^{-2x}} = \frac{\infty}{\infty}$
We can use L'Hospital's Rule:
$\lim\limits_{x \to -\infty}\frac{x^2-x^3}{e^{-2x}}$
$=\lim\limits_{x \to -\infty}\frac{2x-3x^2}{-2e^{-2x}}= \frac{-\infty}{-\infty}$
We can use L'Hospital's Rule:
$\lim\limits_{x \to -\infty}\frac{2x-3x^2}{-2e^{-2x}}$
$=\lim\limits_{x \to -\infty}\frac{2-6x}{4e^{-2x}} = \frac{\infty}{\infty}$
We can use L'Hospital's Rule:
$\lim\limits_{x \to -\infty}\frac{2-6x}{4e^{-2x}}$
$=\lim\limits_{x \to -\infty}\frac{-6}{-8e^{-2x}}$
$= 0$
