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!

This is a bit of a hand-wavy argument but I thought it was fun so I'll share it. Let $f_n(x)=\dfrac{\left(1+x^{\frac{1}{n}}\right)^{n}}{2^{n}}$, and assume $\lim\limits_{n\to\infty}f_n(x)=f(x)$ exists, and that the limit of derivatives, $\lim\limits_{n\to\infty}f_n'(x)$, exists and is equal to $f'(x)$. Then,
$$\begin{align}f_n'(x)&=\underbrace{\frac{(1+x^{\frac{1}{n}})^n}{2^n}}_{f_n(x)}\cdot\frac{x^{\frac{1}{n}}}{x\left(1+x^{\frac{1}{n}}\right)}
\\
\lim_{n\to\infty}f_n'(x)&=\lim_{n\to\infty}f_n(x)\cdot \lim_{n\to\infty}\frac{x^{\frac{1}{n}}}{x\left(1+x^{\frac{1}{n}}\right)}
\\
f'(x)&=\frac{f(x)}{2x}
\\
\int \frac{f'(x)}{f(x)}\ \mathrm{d}x&=\int\frac{1}{2x}\ \mathrm{d}x
\\
\ln |f(x)|&=\frac{1}{2}\ln |x|+C
\\
f(x)&=\pm C_2\sqrt{\left|x\right|}
\end{align}$$
Then, given $f(1)=\lim\limits_{n\to\infty}f_n(1)=\lim\limits_{n\to\infty} 1$ and that $f,x\ge 0$, we see $f(x)=\sqrt{x}$.
