find a limit of $\lim_{n\rightarrow\infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n$ $$\lim_{n\rightarrow\infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n$$
No idea how to do it please help step by step
 A: the natural log of both sides should yield an answer (after applying l'Hopitals rule).
\begin{align*}
\lim_{n \to \infty}f(n)&= \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n \\
\lim_{n \to \infty}\log n&= \frac{\log \left(\frac{a^{1/n}+b^{1/n}}{2}\right)}{1/n} \\
&=\lim_{n \to \infty}\frac{-(a^{1/n}\log(a) +b^{1/n}\log(b))/(2n^2)}{-1/n^2}\\
&=\lim_{n \to \infty}\frac{1}{2}\left(a^{1/n}\log(a) +b^{1/n}\log(b)\right)\\
&=\frac{1}{2}\log(ab)
\end{align*}
So $$\lim_{n \to \infty}f(n)=(ab)^{1/2}$$
A: Assume that there is a $y$ so that
$$
\lim_{n\to\infty}\left(1+\frac{y}{n}\right)^n=x\tag{1}
$$
Then
$$
\lim_{n\to\infty}\left(\frac{1+\frac{y}{n}}{x^{1/n}}\right)^n=1\tag{2}
$$
Since $\frac{1+a}{1+b}$ is between $\frac11$ and $\frac{a}{b}$, $(2)$ implies that
$$
\begin{align}
1
&=\lim_{n\to\infty}\left(\frac{1+1+\frac{y}{n}}{1+x^{1/n}}\right)^n\\[6pt]
&=\lim_{n\to\infty}\left(\frac{1+\frac{y}{2n}}{\frac{1+x^{1/n}}{2}}\right)^n\tag{3}
\end{align}
$$
Squaring $(3)$ and applying $(1)$ gives
$$
\begin{align}
x&=\lim_{n\to\infty}\left(1+\frac{y}{2n}\right)^{2n}\\[6pt]
&=\lim_{n\to\infty}\left(\frac{1+x^{1/n}}{2}\right)^{2n}\tag{4}
\end{align}
$$
Taking the square root of $(4)$ yields
$$
\lim_{n\to\infty}\left(\frac{1+x^{1/n}}{2}\right)^n=\sqrt{x}\tag{5}
$$
Applying $(5)$ to the problem says
$$
\begin{align}
\lim_{n\to\infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n
&=a\lim_{n\to\infty}\left(\frac{1+(b/a)^{1/n}}{2}\right)^n\\
&=a\sqrt{b/a}\\[9pt]
&=\sqrt{ab}\tag{6}
\end{align}
$$
