Answer to Question #91436 in Real Analysis for Kiran

Question

Show that limit n approaches to infinity (xe^-nx)=0 for x € real number , x>0

Answer 1

Let $a = e^{-x}$, $x>0,$ then $a \in (0,1)$. Then It is well-known that

$\lim_{n\to \infty} a^n = 0.$

This fact can be proved by the definition of the limit: in order that $a^n < \varepsilon$ it is sufficient that $n > \log_a \varepsilon$.

Thus,

$\lim_{n\to \infty} xe^{-nx} = x\lim_{n\to \infty} (e^{-x})^n = x \lim_{n\to \infty} a^n = x \cdot 0= 0.$