Find $\lim_{n\rightarrow\infty} \frac{\log(2 + 3^{n})}{2n}$ 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?
 A: $$\frac{\log(2+3^n)}{2n}\sim \frac{\log(3^n)}{2n}=\frac{n\log(3)}{2n}\longrightarrow\frac{\log(3)}{2}\qquad\text{as $n\to\infty$}$$
A: $$\log(2+3^n)\\=\log(3^n(2/3^n+1))\\=\log(3^n)+\log(2/3^n+1),$$
and
$$\frac{\log(3^n)}{2n}=\frac{\log(3)}{2},$$
and
$$\lim_{n\rightarrow+\infty}\frac{\log(2/3^n+1)}{n}=0.$$
A: 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 $\infty$).
A: \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}
A: $$\lim_{n\rightarrow\infty} \frac{log(2 + 3^{n})}{2n}=\lim_{n\rightarrow\infty} \frac{log( 3^{n})+log( \frac{2+3^n}{3^n})}{2n}=\lim_{n\to\infty}\dfrac{n\log 3}{2n}+\lim_{n\to\infty}\dfrac{\log {(1+\frac{2}{3^n})}}{2n}=\frac{\log 3}{2}+\frac{\log 1}{\infty}=\frac{\log 3}{2}$$
A: $$ \frac{\log(2 + 3^{n})}{2n}= \frac{\log(3^n(\frac2{3^n} + 1))}{2n}= \frac{n\log(3)+\log((\frac2{3^n} + 1))}{2n} \to\frac{\log(3)}{2}$$
A: Note that
$$\frac{\log(2 + 3^{n})}{2n}=\frac{\log 3^n+\log(\frac2{3^n} + 1)}{2n}=\frac{n\log 3+\log(\frac2{3^n} + 1)}{2n}\to\frac{\log 3}{2}$$
A: Rewriting the hint properly:
Hint $$log(2+3^{n}) = log(3^{n}) + log(\frac{2 + 3^{n}}{3^{n}})$$
Then your own work leads you to
$$\frac{n log(3)}{2n} + \frac{log(\frac{2 + 3^{n}}{3^{n}})}{2n}$$
Now if I take the limit of this expression:
You get $ \frac{ log(3)}{2} $ as you expected, together with something of the form $\frac{0}{\infty}$, QED.
A: We can generalize a little bit
Let $a_1,$ $\cdots,$ $a_p$ be a  increasing sequence of nonnegative  numbers, $a_p>1$ and $c>0$.
Calculate
$$\lim_{n \to \infty} \frac{\ln \left(a_1^n+ \cdots+a_p^n +c \right)}{n}. $$
We have
$$\frac{1}{n}\ln \left(a_1^n+ \cdots+a_p^n +c \right)=\frac{1}{n}\ln \left(a_p^n \left[\left(\frac{a_1}{a_p}\right)^n+ \cdots+1 +\frac{c}{a_p^n}\right] \right) $$
$$= \frac{n\ln (a_p)}{n} +\frac{1}{n} \ln \left( \left(\frac{a_1}{a_p}\right)^n+ \cdots+1 +\frac{c}{a_p^n} \right)  $$
every term $\left(\frac{a_1}{a_p}\right)^n \to 0$ and the same $\frac{c}{a_p^n} \to 0$, so the above limit by continuity of the $\ln$, is
$$\ln (a_p)=\lim_{n \to \infty}\frac{1}{n} \ln \left(a_1^n+ \cdots+a_p^n +c \right). $$
In this case $a_p=3$ and we multiply by $1/2$ the limit
