Compute $\lim_{n \to \infty}(\frac{a_n+b_n}{2})^n$ I am trying to solve the following problem: 
Compute $\lim_{n \to \infty}(\frac{a_n+b_n}{2})^n$ when $\lim_{n \to \infty} a_n^n=a>0$ and $\lim_{n \to \infty} b_n^n=b>0$ such that $a_n,b_n>0 \ \forall \ n \ \in \mathbb{N}$.
I tried to use the Sandwich Theorem to come up with an answer, but my upper bound was not tight:
$\max(a_n,b_n)\ge(\frac{a_n+b_n}{2}) \ge \sqrt{a_nb_n}$
On passing to the limits I got the following:
$\max(a,b)\ge \lim_{n \to \infty}(\frac{a_n+b_n}{2}) \ge \sqrt{ab}$
But this doesn't help me at all. How could I actually compute the limit?
 A: This is done in steps. First we have to note that both $a_n, b_n$ tend to $1$. This follows from the fact that $n\log a_n\to \log a$ and thus $\log a_n\to 0$.
Next we can let $x_n$ denote the expression whose limit is to be evaluated here. Then we have $$\log x_n=n\log \left(1+\frac{a_n+b_n-2}{2}\right)$$ and the limit of above expression is same as that of $$\frac{1}{2}\cdot\{n(a_n-1)+n(b_n-1)\}$$ Next we can use the fact that $n\log a_n\to\log a$ which implies $n(a_n-1)\to\log a$. The limit of $\log x_n$ is thus equal to $$\frac{\log a +\log b} {2}$$ It follows that $x_n\to\sqrt{ab} $.
The above argument makes use of the standard limit $\lim\limits_{x\to 1}\dfrac{\log x} {x-1}=1$.
A: The key point is that


*

*$a_n^n\to a \implies a_n^n=a+o\left(1\right)\implies a_n=\sqrt[n]a+o\left(\frac1{n}\right)\to 1$

*$b_n^n\to b\implies b_n^n=b+o\left(1\right)\implies b_n=\sqrt[n]b+o\left(\frac1{n}\right)\to 1$
therefore we have that
$$\left(\frac{a_n+b_n}{2}\right)^n
=\left(1+\frac{a_n-1+b_n-1}{2}\right)^n=$$
$$=e^{n\log\left(1+\frac{a_n-1+b_n-1}{2}\right)}\to\sqrt{ab}$$
indeed
$$n\log\left(1+\frac{a_n-1+b_n-1}{2}\right)=
\frac12\frac{a_n-1+b_n-1}{\frac1n}\frac{\log\left(1+\frac{a_n-1+b_n-1}{2}\right)}{\frac{a_n-1+b_n-1}{2}}\to \log\sqrt{ab}$$
since by standard limit $x\to 0,\quad \frac{\log(1+x)}{x}\to 1$
$$\frac{\log\left(1+\frac{a_n-1+b_n-1}{2}\right)}{\frac{a_n-1+b_n-1}{2}}\to
   1$$
and by standard limit $x\to 0,\quad \frac{A^x-1}{x}\to \log A$


*

*$\frac{a_n-1}{\frac1n}=\frac{\sqrt[n]{a_n^n}-1}{\frac1n}=\frac{\left(a+o\left(1\right)\right)^\frac1n-1}{\frac1n}=\frac{a^\frac1n-1}{\frac1n}+o\left(1\right) \to \log a$

*$\frac{b_n-1}{\frac1n}=\dots=\frac{b^\frac1n-1}{\frac1n}+o\left(1\right) \to \log b$
we have
$$\frac12\frac{a_n-1+b_n-1}{\frac1n}=\frac12\left(\frac{a_n-1}{\frac1n}+\frac{b_n-1}{\frac1n}\right)\to \frac12(\log a + \log b)=\log\sqrt{ab}$$
