$\displaystyle a_n= (3^n + 5^n)^{\frac 1n}$
is the above sequence convergent? if so, what is the limit?
Just as an alternative..
If you can use the fact that $\displaystyle \lim_{n \to \infty} a^{\frac{1}{n}} = 1$ for $\displaystyle a > 0$, then notice that:
$\displaystyle 5^n \ \ \leq \ \ 3^n + 5^n \ \ \leq \ \ 5^n + 5^n$
$\displaystyle \Rightarrow \left(5^n \right)^{\frac{1}{n}} \ \ \leq \ \ \left(3^n +5^n \right)^{\frac{1}{n}} \ \ \leq \ \ \left(5^n +5^n \right)^{\frac{1}{n}}$
$\displaystyle \Rightarrow 5 \ \ \leq \ \ \left(3^n +5^n \right)^{\frac{1}{n}} \ \ \leq \ \ 5\left(2 \right)^{\frac{1}{n}}$
And use squeeze theorem ..
recall that, by L'Hopital's, $\displaystyle \lim_{n \to \infty} \frac {\ln (3^n + 5^n)}n = \lim_{n \to \infty} \frac {\frac d{dn}[ \ln (3^n + 5^n)]}{\frac d{dn}n}$
i suppose it is the $\displaystyle \frac d{dn}[ \ln (3^n + 5^n)]$ that is giving you trouble. what did you get for this?