An expression for $\lim_{n\to\infty}\frac1{2^n}\left(1 + x^{1/n}\right)^n$ 
I am looking for a closed form answer to the limit ($x<1$): $$\lim_{n\to\infty}\frac1{2^n}\left(1 + x^{1/n}\right)^n$$


For context, I was studying weighted averages and considered $$\left(\frac12(x^{1/n} + y^{1/n})\right)^{n}$$ to be a good way to weight averages in favour of the lower number (similar to root mean squared, but kinda reversed). I think was studying what happens for different values of $n$. For $x=3$, $y=5$, I found that this limit seems to converge to $3.78962712197\dots$ but I do not recognise where this number comes from. Rearranging the above average formula gives $$\frac{y}{2^n}\left(1+\left(\frac xy\right)^{1/n}\right)^n$$and this is what inspired the question. I see it looks similar to some kind of exponential, but it isn't quite there. I also tried exponentiating and logging the whole expression to bring down the $n$, but I didn't know how to deal with the power inside the brackets then. My main issue is that since $1/n$ goes to $0$, the thing thats raised to this power goes to $1$ (and so is not small) so series expansions can't be used. 
Thanks!

Since the logarithm is continuous, (Interchanging limits and logarithms), with $f=\frac{1}{2^n}(1+x^{1/n})^n$
$$\log \lim_{n\rightarrow \infty} f = \lim_{n\rightarrow \infty}\log f$$
$$\log f = \frac{\log (1+x^{1/n}) - \log 2}{\frac{1}{n}} $$
Applying L'Hopital's rule:
$$\log f \rightarrow \lim_{n\rightarrow \infty}\frac{x^{1/n-1}}{1+x^{1/n}}\log x=\frac{1}{2} \log x $$
Thus 
$$f \rightarrow x^{1/2}$$
