calculus - 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?

Answer

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 $\lim\limits_{x\to 1}\dfrac{\log x} {x-1}=1$.