Computing a limit of sequence

Question

After applying l'Hopital rule twice, one sees that

$$ \lim_{n\to \infty} n a ~e^{-an} =0 \qquad \qquad (a\in [0,1]) . $$ I would like to ask if someone can prove it using different way? Bests.

Answer 1

For any $x>0$, $$ 0 < x e^{-x} = \frac{x}{\left(e^{x/2}\right)^2} < \frac{x}{\left(1+\frac{x}{2}\right)^2}<\frac{4x}{1+x^2}=\frac{4}{x+\frac{1}{x}} $$ hence the claim follows by squeezing. You may also prove: $$ \forall x\geq0,\qquad e^x \geq (1+x)\cdot\left(1+\frac{x^2}{3}\right). $$

Answer 2

Hint for another approach: Let $b_n=an\cdot e^{-an}$, and let $a>0$. Then, $\forall n\ b_n>0$.

But $\lim_{n\to +\infty}{b_{n+1}\over b_n}=e^{-a}<1$, which means...