Evaluating $\lim_{x\to-\infty} xe^{-x}$ Not sure how to think about this one.
$$\lim_{x\to-\infty}  xe^{-x}$$
If I place "$-\infty$" in place for $x$ I get
$$(-\infty)e^{\infty}$$
the limit of which would seem to be $0$ as $x$ approaches $-\infty$.
But in fact the limit is $-\infty$ (if looking at the drawn graph) as $x$ approaches $-\infty$.
 A: Using the product law for limits,
\begin{align}
\lim_{x \to -\infty}xe^{-x} &= \lim_{x \to -\infty}x \cdot \lim_{x \to -\infty}e^{-x} \\
&= -\infty \cdot \infty \\
&= -\infty \, .
\end{align}
Keep in mind though that in limiting analysis, '$\infty$' and '$-\infty$' are both just symbols for unbounded growth. When we write
$$
\lim_{x \to -\infty}e^{-x} = \infty \, ,
$$
all we mean is that $e^{-x}$ gets arbitrarily large as $x$ gets arbitrarily large and negative.
A: Intuitively we can see the limit is $-\infty$:

To make it rigorous we want to find  $x_0(\epsilon)<0$ (a function in terms of $\epsilon<0$) such that
$$x<x_0(\epsilon)\implies xe^{-x}<\epsilon$$
We know that $$e^x \ge1+x \implies e^{-x}\ge 1-x$$So for $x\le 0 $ we have $$xe^{-x}\le x(1-x)$$Let $$x(1-x)\lt \epsilon \implies -x^2 +x - \epsilon \lt 0 \implies x^2 - x+ \epsilon\gt 0$$By choosing $x_0(\epsilon) = \frac{1 - \sqrt{1 - 4\epsilon}}{2} - 1$, the proof is complete. We can check it easily $$x_0(1-x_0) =\epsilon - \sqrt{1 - 4\epsilon} - 1 \lt \epsilon \iff 1 + \sqrt{1 - 4\epsilon} \gt 0$$which is true since $\sqrt{1 - 4\epsilon} \gt 0$.
