# What is the limit of (x^2)(e^x) as x goes to negative infinity?

Oct 8, 2015

${\lim}_{x \rightarrow - \infty} {x}^{2} {e}^{x}$

Attempting to evaluate this limit by simply "plugging in" $- \infty$ will cause the indeterminate form $\infty \cdot 0$

However, if rewrite this as

${\lim}_{x \rightarrow - \infty} {x}^{2} / {e}^{-} x$

We can apply L'Hôpital and go on our merry way

${\lim}_{x \rightarrow - \infty} {x}^{2} / {e}^{-} x = {\lim}_{x \rightarrow - \infty} \frac{2 x}{- {e}^{-} x} = {\lim}_{x \rightarrow - \infty} \frac{2}{e} ^ - x = 0$

Not sure how one would go around doing it without L'Hôpital though.

Oct 8, 2015

$= {\lim}_{x \to - \infty} 2 x {e}^{x} = 0$

#### Explanation:

Use L'Hôpital's Rule:

Since:
${e}^{x} = \frac{1}{{e}^{- x}}$

We can re-write this as:
lim_{x to -infty}x^2e^{x} =lim_{x to -infty}{x^2}/{e^{-x}}

by L'Hôpital's Rule:
$= {\lim}_{x \to - \infty} \frac{2 x}{- {e}^{- x}}$

apply L'Hôpital's Rule again:
$= {\lim}_{x \to - \infty} \frac{2}{{e}^{- x}} = \frac{2}{\infty} = 0$

We can see this by graphing $y = 2 x {e}^{x}$ :

graph{x^2 e^x [-10, 10, -5, 5]}