Find the limit of given function as x approaches infinity

Question

$\ lim_{x \to \infty} x^{4}{e^{-x^{2}}} $. Any tip how to start on this problem? I have no idea how to start on this type of problem.

Answer 1

Simplify with $$x^4e^{-x^2}=4\left(\frac{x^2}2e^{-x^2/2}\right)^2=4\left(te^{-t}\right)^2$$ where $t\ge0$. Then you can solve just for $te^{-t}$.

L'Hospital will work. You can also use $e^t\ge1+t+\dfrac{t^2}2$ from Taylor.

Answer 2

$$\frac{x^4}{e^{x^2}}=\frac{x^4}{1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+...}=2!\frac{x^4}{x^4+2!(1+x^2+\frac{x^6}{3!}+....)}$$ $$\frac{2x^4}{x^4(1+2!(\frac{1}{x^4}+\frac{1}{x^2}+\frac{x^2}{3!}+....)}=\frac{2}{1+2!(\frac{1}{x^4}+\frac{1}{x^2}+\frac{x^2}{3!}+....)}$$ so the limit is $0$