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

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.

• One way I might try is by using the Taylor series expansion of $e^{-x}.$ That way you could divide the denominator of your second expression by $x$, to get \begin{align}\lim _{x \rightarrow -\infty} 1/( 1/x - 1 + x/2 - x^2/6 + \cdots).\end{align}. Taking the limit into the denominator, you could prove that the denominator is infinite and hence that the limit is 0. It's not very rigorous, but I thought I'd throw it out there. – Baylee V Feb 15 at 13:44
• Actually, this should be evaluated with simple calculus theory. – mvfs314 Feb 15 at 13:44
• You're almost there: $\lim_{x\to-\infty}\frac{x}{e^{-x}}=\lim_{x\to-\infty}\frac{1}{-e^{-x}}=\lim_{x\to\infty}\frac{1}{-e^x}$. – SketchySketch Feb 15 at 13:46
• @mvfs314 As you'll see in the answers, there are many ways to approach the question. Can you share with us what you know about the exponential map? How was it defined? What properties you're supposed to know? – mathcounterexamples.net Feb 15 at 14:58
• @mathcounterexamples.net I really can understand all the answers, but this question is for a student who just started to learn the basic of limits. – mvfs314 Feb 15 at 15:11

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.
• Although the inequality is valid and the answer is obviously valid, it still seems like cheating a bit to use an equality directly derived from Taylor's formula, which is much general than L'Hôpital's rule. – PierreCarre Feb 15 at 13:57
• @PierreCarre I strongly disagree. How do you define the exponential? – mathcounterexamples.net Feb 15 at 14:01
• You do not need to define the exponential from its power series representation. You can also define it, for instance, as the solution of $y' = y$, $y(0) = 1$. so it is not necessarily true that you possess a priori the knowledge of the power series. – PierreCarre Feb 15 at 14:03
• @PierreCarre Where have you seen that I used the power series? The inequality $e^x \gt 1 + x +x^2/2$ can be derived from the differential equation. – mathcounterexamples.net Feb 15 at 14:09
• Come on... How did you choose $1+x+\frac 12 x^2$? – PierreCarre Feb 15 at 14:13

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 $$L=0$$.

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,

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