Limit $\lim_{n→∞} [log(2+3^n)]/2n$ Find $\lim_{n→∞} [log(2+3^n)]/2n$
I have my work till the very last step then i dont know how to continue
$\lim_{n→∞} [log(2+3^n)]/2n$
=$\lim_{n→∞} log(3^n)+\lim_{n→∞} log[(2+3^n)/3n]$
=$\lim_{n→∞} log(3^n)+\lim_{n→∞} log[(2/3^n)+1/1]$
="I dont know what it is"$+1$
Since $\lim_{n→∞} log(3^n)$ is infinity???
What should I do?? Thanks!!
 A: I don't see how your first step is valid.
But note
$$
{\log(3^n)\over 2n}\le {\log(2+3^n)\over 2n}\le  {\log( 3^{n+1})\over 2n}.
$$
Now use the Squeeze theorem (the power rule for logarithms will prove useful when computing the required limits of the left and right hand sides of the above inequality).


I think I see now what you were attempting:
$$
{\log(2+3^n)\over2n} 
={\log\bigl(3^n( {2\over3^n}+1)\bigr)\over2n} ={\log 3^n+\log  ( {2\over3^n}+1)
\bigr)\over2n}
=\color{maroon}{{\log 3^n\over 2^n}}+\color{darkgreen}{{\log  ( {2\over3^n}+1)\over 2n}}.
$$
This will prove useful: 
$\displaystyle \qquad\lim\limits_{n\rightarrow\infty}\color{darkgreen}{\log  ( {2\over3^n}+1)\over 2n}=0\ \ $ and  $\displaystyle\ \ \lim\limits_{n\rightarrow\infty}\color{maroon}{\log 3^n\over 2^n}={\log 3\over 2}$;$\ \ $ so
  $\displaystyle\ \ \lim\limits_{n\rightarrow\infty}{\log(2+3^n)\over2n} ={\log 3\over 2}$.
Much cleaner than my solution!
A: Hint: $\lim_{n\rightarrow \infty} \frac { \log (2+3^n)}{\log {3^n} } = 1$
Hint: $\frac {\log 3^n}{2n} = \frac {n \log 3}{2n} = \frac {\log 3}{2}$. Hence the limit (of this sequence to infinity) is $\frac {\log 3}{2}$.
A: Use LHopital's rule. lim log( 2 +3^n)/2n =  lim [ ln(2+3^n)/ln10]/2n
                                       = (1/ln10)lim[(1/(2+3^n))(3^n)ln3]/2
                                       = ln3/2ln10 lim[3^n/(2+3^n)
                                       = ln3/2ln10 lim [(3^n ln3)/(3^n ln3)]
                                       = (1/2)log3
