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.
Are you referring to series or sequences?
- If you're talking about sequences, you'll have in most cases to evaluate its limit for $n\to+\infty$. If for some reasons you can't, e.g. it's given by a recursion formula, then you could use the so-called Cauchy condition: if the sequence lies in a complete space (such as $\mathbb R$, $\mathbb C$), it converges if and only if it satisfies the condition that for all $\epsilon>0$, there exists an $N\in\mathbb N$ such that for all $n, m>N$ you have $|a_n-a_m|<\epsilon$.
- If you're talking about series, directly calculating its limit is not, generally, the easiest way to prove it converges.
There are many other criteria, though, you can use to prove it without calculating the sum.
That explanation you gave with subsequences isn't usually the easiest (at least, as far as I know).
If you can't guess the limit of the whole sequence, you can start with a simpler subsequence: this is just a hunch, though: a sequence converges to a limit $\ell$ if and only if each of its subsequences converges to $\ell$ too.
You can see by yourself that this can't be used to determine the convergence of a sequence: you just can't calculate an infinite amount of limits!
Anyway, if an easy subsequence you found converges to a certain limit, then you know what the overall limit should be. If that's not the case, then your sequence doesn't have a limit.
(Note that this limit I'm speaking of might as well be infinite.)
Here's a not-so-trivial example: $a_n=\cos(\pi n)\frac1{n}$.
Using subsequences, you can try the one in which $\cos(\pi n)$ is constant (it can be either $-1$, $0$ or $1$, as $n$ is an integer), so we'll take $\cos(\pi n)=1$, so $n=2k$ ($k\in\mathbb N$).
Thus we have $a_{2k}=\cos(2k\pi)\frac1{2k}=\frac1{2k}$ which converges to zero.
You should expect, then, that the overal limit of $a_n$ is zero too, but you can't use this result to prove that the limit is really zero (see next example), so you have still to use other methods.
An example of a sequence that doesn't have a limit: the sequence $b_n=(-1)^n$. You can find the subsequence for even values of $n$, that is (using $n=2k$, with $k\in\mathbb N$) $b_{2k}=(-1)^{2k}=1^k=1$, so it converges to $1$.
But then you find another subsequence, for odd $n$ ($n=2k+1$), that is $b_{2k+1}=(-1)^{2k+1}=-1$, so those two have different limits: consequently $b_n$ cannot converge (in fact, it oscillates between $1$ and $-1$).
As for your question «What's the trick to finding the sub-sequence?»: intuition.
Best Answer
We use that $$\log x=\int_1^x t^{-1}dt$$
Let $\alpha>0$; choose $0<\varepsilon <\alpha$. Then $$\frac{\log x}{x^{\alpha}}=\frac{1}{x^{\alpha}}\int_1^x t^{-1}dt<\frac{1}{x^{\alpha}}\int_1^x t^{\varepsilon-1}dt<\frac{x^{\varepsilon-\alpha}}{\varepsilon}\to 0$$