$\lim_{x \to \infty} xe^{-x^2/2} $ without lhospital

Question

I would like to ask you, how to calculate this without l hospital $$\lim_{x \to \infty} xe^{-x^2/2} $$

Answer 1

$e^{u}\geq u$ for $u\geq 0$, so $xe^{-x^{2}/2}\leq\dfrac{x}{2^{-1}x^{2}}=\dfrac{2}{x}\rightarrow 0$.