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
 A: 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 $\lim_{x\to\infty}\frac{x}{2^x}=0$.
A: The limit of form that infinity over infinity can, in most cases, be easily computed using  L'Hospital rule.
Thus $$\lim_{n\rightarrow\infty}\frac{n}{2^n}=\lim_{n\rightarrow\infty}\frac{n'}{(2^n)'}=\lim_{n\rightarrow\infty}\frac{1}{2^n\ln 2}=0$$
A: The term  indeterminate forms applies in your case only if you want to calculate the limit of a sequence of quotients as a quotient of limits:
$$\frac{\lim \limits_{n \rightarrow \infty}{n}}{\lim \limits_{n \rightarrow \infty}{2^n}}$$
But this view is not necessary, because you just have a sequence of rational numbers and you can directly apply the $\epsilon$ definition of a limit to your sequence: Given $\epsilon > 0$, you have $0\lt \frac{n}{2^n} \le \frac{2}{n-1} < \epsilon$ for $n > \frac{2}{\epsilon} + 1$, this shows that the sequence converges to 0. 
