Limit of $\frac{2^{\frac{\ln n}{2}}}{n}$ I would like to know the limit of $\lim_{n\to\infty}\frac{2^{\frac{\ln n}{2}}}{n}$. I notice repeatedly taking L'opital Rule does not help. Is there any other approach where I can use to derive the limit?
 A: Hint: $$2^{(\ln n)/2}=2^{(\log_2 n)\frac{\ln 2}{2}}=\left(2^{(\log_2 n)}\right)^{(\ln 2)/2}=n^{(\ln 2)/2}$$
Now simplify further, using the laws of exponents on numerator and denominator of the original expression.
A: Let 
$$y=\dfrac1n2^{\frac12\ln n}\Rightarrow \ln y=\ln n\left(\dfrac{\ln 2}{2}-1\right)\to-\infty$$
as $n\to\infty$, then $y\to0$.
A: I like @vadim123's answer, but personally I find coming up with the algebra a little more intuitive if I explicitly include the base of the logarithm I'm trying to undo. We have a $\ln n$ thing going on here, so we want to incorporate $e$ somehow.
$$2^{(\ln n)/2}=\left(e^{\ln 2}\right)^{(\ln n)/2}=e^{(\ln n)(\ln 2)/2}=n^{(\ln 2)/2}$$
Just for fun there's a completely different trick you can do still using L'Hopital's rule. Note that this only computes the limit if it exists; proof of existence has to be taken care of separately.
When you make no simplifications to the top or the bottom and apply L'Hopital's rule, you get $$\lim_{n\to\infty}\frac{2^{\frac{\ln n}{2}}}{n}=\lim_{n\to\infty}\frac{\frac{\ln 2}{2}2^{\frac{\ln n}{2}}}{n}.$$
Pulling the constant $\frac{\ln 2}{2}$ out and calling the limit you want computed $x$, we find $$x=\frac{\ln 2}{2}x.$$ Solving for $x$, the limit is $0$.
A: Set $e^y =2$, then  $y \in (0,1)$ (why?).
Rewriting:
$2^{((\log n)/2)}=e^{(y/2)(\log n)}=$
$e^{\log n^{y/2}}= n^{y/2}$, $y/2 \in (0,1).$
Now consider the limit $n \rightarrow \infty$ of $\dfrac{n^{y/2}}{n}$.
