Find $$\lim_{n\rightarrow\infty} \frac{\log(2 + 3^{n})}{2n}$$
Hint: $$\log(2+3^{n}) = log(3^{n}) + \log(\frac{(2 + 3^{n})}{3^{n}})$$
Attempt:
If I apply the hint to the expression and do a little simplification I arrive at:
$$\frac{n log(3)}{2n} + \frac{\log(\frac{log(2 + 3^{n})}{3^{n}})}{2n}$$
Now if I take the limit of this expression:
$$\lim_{n\rightarrow\infty} \frac{ \log(3)}{2} + \lim_{n\rightarrow\infty} \frac{\log(\frac{\log(2 + 3^{n})}{3^{n}})}{2n}$$
Here is where I am stuck. I feel I can argue that the limit of the second term goes to 0 because the denominator will go to infinity faster than the numerator. As such I am left with $\frac{log(3)}{2}$ as my answer. My only concern is that I could've used that same train of thought with the original expression.
Advice?