Find $\lim_{n\rightarrow\infty} \frac{\log(2 + 3^{n})}{2n}$

$$\frac{\log(2+3^n)}{2n}\sim \frac{\log(3^n)}{2n}=\frac{n\log(3)}{2n}\longrightarrow\frac{\log(3)}{2}\qquad\text{as $n\to\infty$}$$

The hint is not quite right. $log(2+3^n)=log(3^n)+log(1+\frac{2}{3^n})$. Using this and the fact that $log(3^n)=n~log(3)$ you get $\frac{log(2+3^n)}{2n}=\frac{log(3)}{2}+\frac{log(1+\frac{2}{3^n})}{2n}$ which tends to $\frac{log(3)}{2}$ as $\frac{log(1+\frac{2}{3^n})}{2n}$ tends to $0$ (the top goes to $log(1)=0$ and the bottom to $\infty$).

$log(2+3^n)=log(3^n(2/3^n+1)=log(3^n)+log(2/3^n+1)$, $log(3^n)/2n=log(3)/2$, $lim_{n\rightarrow+\infty}log(2/3^n+1)/n=0$.

\begin{eqnarray}\lim_{n\rightarrow\infty} \frac{\log(2 + 3^{n})}{2n} &=&\lim_{n\rightarrow\infty} \frac{\log(2/3^n + 1)+\log3^n}{2n}\\\\&=&\lim_{n\rightarrow\infty} \frac{{1\over n}\log(2/3^n + 1)+\log 3}{2}\\&=&{\log 3\over 2}\end{eqnarray}

$$\lim_{n\rightarrow\infty} \frac{log(2 + 3^{n})}{2n}=\lim_{n\rightarrow\infty} \frac{log( 3^{n})+log( \frac{2+3^n}{3^n})}{2n}=\lim_{n\to\infty}\dfrac{n\log 3}{2n}+\lim_{n\to\infty}\dfrac{\log {(1+\frac{2}{3^n})}}{2n}=\frac{\log 3}{2}+\frac{\log 1}{\infty}=\frac{\log 3}{2}$$