Question

lim x approches infinity x^2 e^-x

Answer 1

solve it

Answer 2

\[\lim_{x\to\infty}\frac{x^2}{e^x}=\frac{\infty}{\infty}\]
Apply L'Hopital's rule:
\[\lim_{x\to\infty}\frac{2x}{e^x}=\frac{\infty}{\infty}\]
And once more:
\[\lim_{x\to\infty}\frac{2}{e^x}=\cdots \]

Answer 3

Alternatively, you can use the series representation of \(e^x\) to rewrite:
\[\lim_{x\to\infty}\frac{x^2}{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots}\]
\[\lim_{x\to\infty}\frac{x^2}{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots}\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}\]
\[\lim_{x\to\infty}\frac{1}{\frac{1}{x^2}+\frac{1}{x}+\frac{1}{2!}+\frac{x}{3!}+\frac{x^2}{4!}+\frac{x^3}{5!}+\cdots}\]