real analysis - finding $lim_{n rightarrow +infty}frac{n}{2^n}= ?$

Finding the limit below..:

$$\lim_{n \rightarrow +\infty}\frac{n}{2^n}= ?$$

I really think its 0. But intuitively, infinity over infinity. how can that be? indeterminate forms? Thanks

Answer

Intuitively, $2^n$ grows much faster than $n$.

Note that by the Binomial Theorem, $2^n=(1+1)^n=1+n+\frac{n(n-1)}{2}+\cdots$.

In particular, if $n\gt 1$, we have $2^n\ge \dfrac{n(n-1)}{2}$.

Thus $0\lt \dfrac{n}{2^n}\le \dfrac{2}{n-1}$. But $\frac{2}{n-1}$ approaches $0$ as $n\to\infty$, so by Squeezing, so does $\dfrac{n}{2^n}$.

Another way: You can use L'Hospital's Rule to show $\lim_{x\to\infty}\frac{x}{2^x}=0$.