Prove that limit of $ne^{-an}$ as $n$ goes to infinity equals $0$

Question

Somehow I learnt that the limit of $ne^{-an}$ (for $a > 0$) is $0$ as $n$ goes to infinity. I was trying to convert the expression to the form $\frac{e^{-an}}{\frac{1}{n}}$ and then use the L'Hôpital's rule but then I am getting the symbol $0\cdot \infty$ over and over again. How can prove it?

Answer 1

The original post had a tag "No L'Hospital's Rule." The ensuing presentation was written under that constraint.

In THIS ANSWER, I showed using only (i) the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality

$$e^x\ge \left(1+\frac xn\right)^n$$

for $x>-n$. We can set $n=2$ and assert that for $x>-2$,

$$e^x\ge 1+x+\frac14 x^2$$

Then, we have

$$\begin{align} 0&< ne^{-an}\\\\ &=\frac{n}{e^{an}}\\\\ &\le\frac{n}{1+an+\frac14 (an)^2} \end{align}$$

whereupon applying the squeeze theorem yield the expected limit!