Finding $\lim_{x\to-\infty}x\cdot e^x$ without using L'Hospital rule

Question

I have to evaluate $$\lim_{x\to-\infty}x\cdot e^{x}$$ I know how I could do using L'Hospital rule, but I can't use it. I tried to rewrite it as: $$\lim_{x\to-\infty}\frac{x}{e^{-x}}$$ but I still get $-\infty/\infty$ and I don't what to do next.

Answer 1

For $x \gt 0$ you have $e^x \gt 1 + x + x^2/2 \gt x^2/2$ and therefore for $x \lt 0$ $e^{-x} \gt x^2/2$. Hence for $x \lt 0$

$$0 \lt \vert x \cdot e^x \vert=\frac{\vert x \vert}{e^{-x}} \lt \frac{2}{\vert x \vert}$$ which allows to conclude with the squeeze theorem.

Note (following comments discussion): in this kind of question, the definition (or properties) that you use for the exponential map is critical.

The inequality $e^x \gt 1 + x + x^2/2 \gt x^2/2$ can be derived in several ways:

• Using $e^x= 1 + x + \frac{x^2}{2} + \dots$ Taylor series.
• Using the fact that $\left(e^x\right)^\prime = e^x$ and $e^0=1$.
• Using growth properties like $e^x \gt x^n$ for any $n \in \mathbb N$ and $x$ large enough.

Answer 2

Let us denote $$f(x)=x \mathrm{e}^x .$$

The derivative is $$f'(x)=(x+1)\mathrm{e}^x, $$ and therefore the function is decreasing in $(-\infty ,-1)$. In addition, $f(x)<0$ on that interval, so the monotone convergence theorem gives the existence of $$L=\lim_{x \to -\infty} f(x) \in (-\infty , 0 ]. $$

On the other hand $$\int_{-\infty}^0 f(x) \mathrm{d} x = f(x)-\mathrm{e}^x \big|^0_{-\infty} = -1 - \lim_{x \to -\infty} f(x) + \lim_{x\to -\infty} \mathrm{e}^x = -1-L. $$ As the improper integral converges, we must have $L=0$.

Answer 3

Even though there are already several fine answers, I think the instructor probably just wanted you to use some "notable limit" out of a list. A probable candidate would be $$ \lim_{x\to +\infty} \frac{e^x}{x^p} = +\infty. $$

in this case,

$$ \lim_{x\to -\infty}x e^x = \lim_{y\to +\infty} -y e^{-y} = \dfrac{-1}{\displaystyle \lim_{y\to +\infty} \frac{e^y}{y}} = \frac{-1}{+\infty} = 0. $$