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Determine whether the sequence converges or diverges. If it converges, find the limit.

$$\left\{\frac{\ln4n}{\ln12n}\right\}$$

$\lim_{n\to \infty}\frac{\ln4n}{\ln12n} =$ ?

I am having trouble with this problem, any help is appreciated!!

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the limit is $1$. Write:$$\dfrac{\ln (4n)}{\ln (12n)} = \dfrac{\ln 4 + \ln n}{\ln 12 + \ln n} = \dfrac{1+\dfrac{\ln 4}{\ln n}}{1+\dfrac{\ln 12}{\ln n}}$$.

If instead you want to change the question to: $\displaystyle \lim_{n\to \infty} \dfrac{\ln(4+n)}{\ln(12+n)} = \text{ ?? }$, then you would apply the same trick as above:

$\ln(4+n) = \ln n + \ln\left(1+\dfrac{4}{n}\right)$, and divide top and bottom by $\ln n$ again as above.

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  • $\begingroup$ I have a quick questions about why you separated it from ln(4n) to ln4 + lnn and on the bottom as well? and what would happen if you started out with ln(4+n) instead of ln(4n) on the top? $\endgroup$ – Erick Mar 23 '15 at 4:36
  • $\begingroup$ I edited it, check it out and ask if you have further question about it. $\endgroup$ – DeepSea Mar 23 '15 at 5:22
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Similar to BRICS-Fan but a little simpler:

$\begin{array}\\ \dfrac{\ln4n}{\ln12n} &=\dfrac{\ln n+\ln 4}{\ln n+\ln 12}\\ &=\dfrac{\ln n+\ln 12-\ln 12+\ln 4}{\ln n+\ln 12}\\ &=1+\dfrac{-\ln 12+\ln 4}{\ln n+\ln 12}\\ &=1+\dfrac{-\ln 3}{\ln n+\ln 12}\\ &\to 1\\ \end{array} $

since $\frac{-\ln 3}{\ln n+\ln 12} \to 0 $.

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