lim n→∞ ((n√a + n√b)/2)^n

Explanation:

Step1: Use the fact that $\lim_{n\rightarrow\infty}n(\sqrt[n]{x} - 1)=\ln x$

Let $a_n=\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n$. Take the natural - logarithm of $a_n$: $\ln a_n=n\ln\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)$. Since $\sqrt[n]{a}=e^{\frac{\ln a}{n}} = 1+\frac{\ln a}{n}+o\left(\frac{1}{n}\right)$ and $\sqrt[n]{b}=e^{\frac{\ln b}{n}}=1 + \frac{\ln b}{n}+o\left(\frac{1}{n}\right)$ as $n\rightarrow\infty$. Then $\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}=\frac{(1 + \frac{\ln a}{n}+o(\frac{1}{n}))+(1+\frac{\ln b}{n}+o(\frac{1}{n}))}{2}=1+\frac{\ln a+\ln b}{2n}+o\left(\frac{1}{n}\right)$.

Step2: Use the property $\ln(1 + x)=x+o(x)$ as $x\rightarrow0$

We know that $\ln a_n=n\ln\left(1+\frac{\ln a+\ln b}{2n}+o\left(\frac{1}{n}\right)\right)$. Since $\ln(1 + x)=x+o(x)$ as $x\rightarrow0$, when $x = \frac{\ln a+\ln b}{2n}+o\left(\frac{1}{n}\right)$, we have $\ln a_n=n\left(\frac{\ln a+\ln b}{2n}+o\left(\frac{1}{n}\right)\right)=\frac{\ln a+\ln b}{2}+o(1)$.

Step3: Find the limit of $\ln a_n$ and then $a_n$

Take the limit as $n\rightarrow\infty$ of $\ln a_n$. $\lim_{n\rightarrow\infty}\ln a_n=\frac{\ln a+\ln b}{2}=\ln\sqrt{ab}$. Then, by the continuity of the exponential function, $\lim_{n\rightarrow\infty}a_n = e^{\lim_{n\rightarrow\infty}\ln a_n}=\sqrt{ab}$.

Answer:

$\sqrt{ab}$