how to find the limit $ \lim_{x\to -\infty} xe^x = 0 $ using only simple algebraic manipulation

Question

I found the following limit in an exercise book: $$ \lim_{x\to -\infty} xe^x = 0 $$ this limit is supposed to be solvable without knowledge of L'hospital nor more advanced techniques such as Taylor or others.
So the question is, how can I find that this limit is $ 0 $ using only simple algebraic manipulation or the common limit list?

Answer 1

First a substitution: set $t=-x\;(t\to+\infty)$. You then have to find the limit of $\;-\dfrac t{\mathrm e^ t}$ when $t\to+\infty$. It's a high school result that this limit is $0$.

However, here is the standard proof in high school:

Taking the log, it amounts to proving that $\lim_{t\to +\infty}(\ln t-t)=-\infty$.

For this, you need a lemma:

For all $t\ge 1$, one has $\;\ln t < 2\sqrt t$.

This is an easy consequence of the inequality on derivatives $$\frac 1t <\frac 1{\sqrt {t\mkern2mu}}$$ and the fact that $\ln 1 < 2\sqrt 1$.

Using this lemma, we have, for all $t>1$: $$\ln t-t < 2\sqrt t-t\to -\infty\enspace\text{when}\enspace t\to +\infty.$$

Answer 2

$\lim_{x\to -\infty} xe^x= \lim_{x\to \infty}(-x)\frac{1}{e^x}$, hence we investigate

$\lim_{x\to \infty}x \frac{1}{e^x}$.

For $x>0$ we have $e^x \ge \frac{x^2}{2}$ (power series expansion). Thus

$ 0 \le \frac{x}{e^x} \le \frac{2}{x}$. This gives: $\lim_{x\to \infty}x \frac{1}{e^x}=0$.

Therefore $\lim_{x\to -\infty} xe^x=0$.

Answer 3

Let $x=-y\to +\infty$

$$xe^x=-\frac{y}{e^y}\to 0$$

by squeeze theorem. Indeed since by exponential series $e^y>\frac{y^2}2$

$$0<\frac{y}{e^y}<\frac{2y}{y^2}=\frac2y\to 0$$