How to solve this limit $\lim_{n\to \infty}(\frac{2^n}{n^k})$? It's not allowed to apply L'Hopital's rule .
I tried using binomial distribution $2^n=(1+1)^n$ in order to manipulate with numerator but don't know what to do with denominator.
P.s Should i use Squeeze Theorem to this kind of limit .
I need help with this Calculus I problem ...
 A: Consider the sequence $a_n=\frac{2^n}{n^k}$.
$$\frac {a_{n+1}}{a_n}= 2\frac{n^k}{(n+1)^k}=2\left(\frac{n}{n+1} \right)^k =2\left(1-\frac{1}{n+1} \right )^k.$$
Since $k$ is fixed, for large $n$ this ratio exceeds, say $\frac 32$, so $\lim a_n = \infty$.
A: Take the logarithm of the expression (assuming $k \ge 1$)
$$
\ln\left(\frac{2^n}{n^k}\right) =\ln(2^n)-\ln(n^k) = n\ln(2) +\ln\left(\frac{1}{n^k} \right)
$$
It follows
\begin{align}
\lim_{n\to \infty}\ln\left(\frac{2^n}{n^k}\right)
&= \lim_{n\to \infty}\left[n\ln(2) +\ln\left(\frac{1}{n^k} \right)\right] \\
&=\lim_{n\to \infty}(n\ln(2)) + \lim_{n\to \infty}\ln\left(\frac{1}{n^k} \right) \\
&=+\infty + 0 = +\infty
\end{align}
So the limit of the logarithm is $+\infty$ and so is the limit
$$
\lim_{n\to \infty}\left(\frac{2^n}{n^k}\right) = +\infty.
$$
Simply explained: $\ln(x) \to +\infty \implies x \to +\infty$
A: Suppose $k>0$. Let $m=[k]+1$. Then, for $n>2m+2$
$$ \frac{2^n}{n^k}=\frac{(1+1)^n}{n^k}\ge \frac{\binom{n}{m}}{n^{m+1}}=\frac{n(n-1)\cdots(n-m+1)}{m!n^{m+1}}\ge \frac{(n-m+1)^{n-m}}{m!n^{m+1}}=\frac{1}{m!}(1-\frac{m}{n})^{m+1}(n-m)^{n-2m-1} $$
Since
$$\lim_{n\to \infty}\frac{1}{m!}(1-\frac{m}{n})^{m+1}(n-m)^{n-2m-1}=\infty $$
hence one has
$$ \lim_{n\to \infty}\frac{2^n}{n^k}=\infty. $$
A: Answer :
$$\lim_{n\to+\infty} \frac{2^n}{n^k}=\lim_{n\to+\infty}\frac{e^{n\ln(2)}}{e^{k\ln(n)}}=  \lim_{n\to+\infty}e^{n(\ln(2)-k\frac{\ln(n)}{n}) } 
$$
Note that
$$
\lim _{n\to+\infty} k\frac{\ln(n)} {n} =0$$
So :
$$\lim_{n\to+\infty} \frac{2^n}{n^k}=\lim_{n\to+\infty} e^{n\ln(2)}=+\infty $$
Because $\ln(2)>0$.
