Limit of $x^2e^x$as $x$ approaches negative infinity without using L'hopital's rule

I'm trying to find the limit of $e^x x^2$ as $x$ approaches negative infinity:

$$\lim_{x\to-\infty}e^xx^2$$

without using L'hopital's rule. Any suggestions?

• Can you use taylor's series? – 5xum Apr 13 '15 at 6:16
• preferably not, i'm trying to find a way to do this using only basic limit laws (addition, subtraction, multiplication, powers etc) or the squeeze theorem – Timothy Apr 13 '15 at 6:18

Use $$\tag1e^t\ge 1+t\qquad\text{for t\in\mathbb R},$$ possibly the single most useful inequality about the exponential funciton. Then for $t\ge 0$, $e^{3t}=(e^t)^3\ge (1+t)^3\ge t^3$ and for negative $x$ $$0<e^xx^2=\frac{x^2}{e^{3\cdot (-x/3)}}\le\frac{x^2}{(-x/3)^3}=\frac{27}{|x|}$$
• When $x<0$, that's the same @Timothy – user99914 Apr 13 '15 at 6:42
• Simple: $-x=|x|$ for $x<0$ – gst Apr 13 '15 at 6:44
$$\lim_{x\rightarrow -\infty} x^2e^x\\y=-x\\so \\\lim_{y\rightarrow +\infty} (-y)^2e^{(-y)}=\\\lim_{y\rightarrow +\infty} \frac{y^2}{e^y}=?$$ as you know :$e^y=1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+...$ now see $$\lim_{y\rightarrow +\infty} \frac{y^2}{e^y}= \\lim_{y\rightarrow +\infty} \frac{y^2}{1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+...}$$ now $$0 \leq \frac{y^2}{1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+...} \leq \frac{y^2}{\frac{y^3}{3!}+\frac{y^4}{4!}+...}=\frac{y^2}{y^3(\frac{1}{3!}+\frac{y}{4!}+...)} \rightarrow 0$$ so by squeeze theorem ,this lim goes to zero