# Find the limit of the sequence containing logarithm??

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!!

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I dont know how it will work on this but another thing I tried to make it = $\lim_{n→∞} nlog(3)$ = $log(3)\lim_{n→∞} n$ still = ∞ –  Paul Feb 2 at 21:24
How did you get your first step?? –  Calvin Lin Feb 2 at 21:41

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!

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I dont quite familiar with Squeeze theorem. Isn't it only useful when left and right limit are equal? –  Paul Feb 2 at 21:27
@Paul Compute $\lim\limits_{n\rightarrow\infty}{\log(3^n)\over 2n }$ and $\lim\limits_{n\rightarrow\infty}{\log(3^{n+1})\over 2n }$. You'll see they are equal. This implies that $\lim\limits_{n\rightarrow\infty}{\log(2+3^n)\over 2n }$ exists and is equal to this common value. –  David Mitra Feb 2 at 21:30
So its $\lim_{n→∞} log(3n)/2n = \lim_{n→∞} nlog3 / 2n = \lim_{n→∞} log3 / 2 = log3 / 2$ ? –  Paul Feb 2 at 21:35
@Paul Yes, that's correct. –  David Mitra Feb 2 at 21:36
but i still dont know how you come up with this squeeze inequality... like how would i even think of that when doing the problems.. –  Paul Feb 2 at 21:37

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 $\frac {\log 3}{2}$.

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