$\lim _{x\to -\infty }\left(xe^x\right)$ without the use of L'Hospital's rule

Question

I know that if $x<0$, $$\lim _{x\to -\infty }\left(xe^x\right)=\lim _{x\to -\infty }\left(\frac{x}{\frac{1}{e^x}}\right)=[*]$$

With Hospital's rule I will have

$$[*]=\lim _{x\to -\infty }\left(\frac{1}{-e^{-x}}\right)=\frac 1{-\infty}=0$$

What is the easiest best way to have a trick for high school students without the use of Hospital's rule?

Answer 1

It's pretty clear from the graph that $x^2 \leq e^x $ for $x$ large enough. Therefore,

$$0 \leq \lim\limits_{x \to -\infty}|x e^x| = \lim\limits_{x \to \infty} \frac{x}{e^x} \leq \lim\limits_{x \to \infty} \frac{x}{x^2} = 0.$$

This shows that $\lim\limits_{x \to -\infty}|x e^x|$ and hence $\lim\limits_{x \to -\infty}x e^x$ is zero.

Answer 2

Proving not using L'Hospital's rule, this is the simplest I can think of.

Choose $x=-3t\ln t$ for $t>1$, we have

$$0\ge xe^x=-3t\ln t e^{-3t\ln t}\ge-3t\ln t e^{-3\ln t}$$$$=-3t\ln t/t^3\ge -3\ln t/t^2\ge-3(t-1)/t^2\ge -3/t$$

Thus, when $t\to\infty$, $xe^x$ is sandwiched between $-3/t$ and $0$. Thus, it goes to zero.

What we have used, possibly not that "high school", is $\ln x\le x-1$.

Answer 3

Let consider $f(x)=xe^x$

We have $f'(x)=e^x(x+1)<0$ for $x<-1$

So $f$ is continuous, negative and decreasing on $(-\infty,-1]$ and therefore bounded on this interval.

By the addition formula we get $|f(2x)|=|2x\,e^{2x}|=\underbrace{2f(x)^2}_\text{bounded}\times\underbrace{\dfrac 1{|x|}}_{\to 0}\to 0$ at minus infinity.