What is $\lim\limits_{n\to\infty}\frac{n^\sqrt n}{2^n}$? I am stuck at this question where I have to calculate what is big O of
$2^n $and $n^\sqrt{n}$
Can I say that lim $2^n/n^\sqrt{n}$ = $\lim_{n\to\infty} (2/n^{1/\sqrt{n}})^n$
and then conclude that when it means $(2/0)^n\to \infty$ ?
Any help would be appreciated
 A: \begin{align*}
\lim_{n\rightarrow \infty }\frac{n^{\sqrt{n}}}{2^{n}}&=\lim_{n\rightarrow \infty }\exp\left ( \sqrt{n}\ln n-n\ln2 \right )\\
&=\exp\lim_{n\rightarrow  \infty }\left (  \sqrt{n}\left (  \ln n-\sqrt{n}\ln2\right )\right )\\
&=\exp\lim_{n\rightarrow \infty }\sqrt{n}\cdot \ln\left (  \frac{n}{2^{\sqrt{n}}}\right )
\end{align*}
when $n\to \infty$, $\ln\left (  \dfrac{n}{2^{\sqrt{n}}}\right )\to -\infty
$, so $\displaystyle \lim_{n\rightarrow \infty }\sqrt{n}\cdot \ln\left (  \frac{n}{2^{\sqrt{n}}}\right )\to -\infty$
, hence 
$$\lim_{n\rightarrow \infty }\frac{n^{\sqrt{n}}}{2^{n}}=0$$
A: It's not true that $n^{1/\sqrt{n}}\rightarrow 0$. It actually goes to $1$ since you can write it $$e^{\log(n)/\sqrt{n}}$$ and the exponent goes to $0$. 
But the conclusion that $(2/1)^n\rightarrow\infty$ holds (so your original limit, which was the reciprocal, is zero).
A: Consider the sequence defined by :
$$u_n=\frac{n^\sqrt n}{2^n}$$
and then compute :
$$\ln(u_n)=\sqrt n\ln(n)-n\ln(2)$$
Factoring by the leading term :
$$\ln(u_n)=n\left(\frac{\ln(n)}{\sqrt n}-\ln(2)\right)$$
It is well known that $\lim_{t\to\infty}\frac{\ln(t)}{t}=0$; so :
$$\frac{\ln(n)}{\sqrt n}=2\frac{\ln(\sqrt n)}{\sqrt n}\underset{n\to\infty}{\longrightarrow}0$$
hence :
$$\lim_{n\to\infty}\ln(u_n)=-\infty$$
and finally, by continuity of the exponential function :
$$\boxed{\lim_{n\to\infty}u_n=0}$$
A: You can also write $2^n = \left(2^{\sqrt{n}}\right)^{\sqrt{n}}\implies a_n = \left(\dfrac{n}{2^{\sqrt{n}}}\right)^{\sqrt{n}}$, and you can show $2^{\sqrt{n}} > (\sqrt{n})^3=n\sqrt{n}\implies \dfrac{n}{2^{\sqrt{n}}} < \dfrac{1}{\sqrt{n}}$, and the answer of $0$ will follow.
A: $${n^{\sqrt n}\over 2^n}=e^{\sqrt n\ln n-n\ln2}$$
so the behavior of the sequence boils down to the behavior of $\sqrt n\ln n-n\ln2$.  It's convenient to replace $n$ with $n^2$, which gives $2n\ln n-n^2\ln2$.  Since $\ln n\lt{1\over4}n$ for $n\gt$ something, we have $2n\ln n-n^2\ln2\lt({1\over2}-\ln2)n^2\to-\infty$ as $n\to\infty$, since ${1\over2}\lt\ln2$.  It follows that $n^{\sqrt n}/2^n\to0$.
A: Write it as $$2^{\sqrt{n}\log_2 n-n}$$
Now, when $2^{2k}\leq n<2^{2k+2}$ then $2k\leq \log_2 n <2k+2$ and thus $$\sqrt{n}\log_2 n -n <2^{k+2}(k+1)-2^{2k}$$
But for $k\geq 6$, $k+1<2^{k-3}$. So:
$$\sqrt{n}\log 2 - n < 2^{2k-1}-2^{2k}=-2^{2k-1}$$
So $\lim (\sqrt{n}\log_2 n -n) =-\infty$ and the limit you want is $0$.
