Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 40

Determine the $\displaystyle \lim_{x \to -\infty} x^2 e^x$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

$\displaystyle \lim_{x \to -\infty} x^2 e^x = \lim_{x \to -\infty} \frac{x^2}{e^{-x}}$

By applying L'Hospital's Rule...
$\displaystyle \lim_{x \to -\infty} \frac{-x^2}{e^{-x}} = \lim_{x \to -\infty} \left( \frac{2x}{e^{-x}(-1)} \right) = \lim_{x \to -\infty} \left( - \frac{2x}{e^{-x}} \right)$

We will still get an indeterminate form if we evaluate the limit, so we must use the L'Hospital's Rule once more... Thus,

$
\begin{equation}
\begin{aligned}
\lim_{x \to -\infty} \left( -\frac{2x}{e^{-x}} \right) &= \lim_{x \to -\infty} \left( -\frac{2(1)}{e^{-x}(-1)} \right)\\
\\
&= \lim_{x \to -\infty} \frac{2}{e^{-x}}\\
\\
&= \frac{2}{e^{-(-\infty)}}\\
\\
&= \frac{2}{e^{\infty}}\\
\\
&= \frac{2}{\infty}\\
\\
&= 0
\end{aligned}
\end{equation}
$