$\lim _{x\to -\infty }\left(xe^x\right)$ without the use of L'Hospital's rule 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?
 A: 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.
A: 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.
A: 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 $\ln x\le x-1$.
