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

Question

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

Hint: $$\log(2+3^{n}) = \log(3^{n}) + \log\left(\frac{(2 + 3^{n})}{3^{n}}\right)$$

Attempt:

If I apply the hint to the expression and do a little simplification I arrive at:

$$\frac{n \log(3)}{2n} + \frac{\log\left(\frac{\log(2 + 3^{n})}{3^{n}}\right)}{2n}$$

Now if I take the limit of this expression:

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

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?

Answer 1

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

Answer 2

\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}

Answer 3

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

Answer 4

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

Answer 5

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 $\infty$).

Answer 6

Rewriting the hint properly:

Hint $$\log(2+3^{n}) = \log(3^{n}) + \log\left(\frac{2 + 3^{n}}{3^{n}}\right)$$

Then your own work leads you to $$\frac{n \log(3)}{2n} + \frac{\log\left(\frac{2 + 3^{n}}{3^{n}}\right)}{2n}$$

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.

Answer 7

$$ \frac{\log(2 + 3^{n})}{2n}= \frac{\log(3^n(\frac2{3^n} + 1))}{2n}= \frac{n\log(3)+\log((\frac2{3^n} + 1))}{2n} \to\frac{\log(3)}{2}$$

Answer 8

Note that

$$\frac{\log(2 + 3^{n})}{2n}=\frac{\log 3^n+\log(\frac2{3^n} + 1)}{2n}=\frac{n\log 3+\log(\frac2{3^n} + 1)}{2n}\to\frac{\log 3}{2}$$

Answer 9

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