<s>[INST] 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?
 [/INST] $$\lim_{x\rightarrow -\infty} x^2e^x\\y=-x\\so \\\lim_{y\rightarrow +\infty} (-y)^2e^{(-y)}=\\\lim_{y\rightarrow +\infty} \frac{y^2}{e^y}=?$$ 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 
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