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Let $$a_n = \frac{n^3-5n^2+13n-5}{24n^3+3n+1}.$$

Proof whether this sequence converges. So I think limit is $0$, so I've got to prove that this sequence is a null-sequence. But how do I find an $N$ now so that all of the so that for all $n$, $|a_n|<\epsilon$.

Edit: the limit is $1/24$.

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    The limit is $\;\cfrac1{24}\;$ .... – 2017-01-03
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    Thanks, I will edit my post – 2017-01-03
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    **Hint:** Divide both the denominator and the numerator by $n^3$ and then apply the algebra of limits of sequences. – 2017-01-03

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Intuitively, $n^3$ is much bigger than $n^2$ and $n$ (as $n$ tends to infinity). So you may think that the numerator grows as $n^3$ and forget about the rest, and the denominator grows as $24\ n^3$ ($24$ times more).

To capture that intuition, take common factor $n^3$ in both the numerator and the denominator and see what happens. If you get stuck let me know.

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Let $\epsilon>0$.

Note $\left|\frac{n^3-5n^2+13n-5}{24n^3+3n+1}-\frac{1}{24}\right|<\epsilon$ and find $N(\epsilon)$, for which it holds for all $n>N$.

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The trick is to break out the limit and the $n^3$ terms:

$${n^3 - 5n^2 + 13n -5 \over 24n^3 + 3n + 1} = {n^3(1-5/n + 13/n^2 - 5/n^3)\over 24n^3(1+3/24n+1/24n^3)} = {1\over24}{(1-5/n + 13/n^2 - 5/n^3)\over(1+3/24n+1/24n^3)}$$

Now you only have to estimate $1-5/n+13/n^2-5/n^3$ and $1+3/24n+1/24n^3$ for large $n$ and make sure they are suficiently near $1$. We have:

$$1 > 1 - 5/n + 13/n^2 - 5/n^3 > 1-5/n - 5/n^3 > 1-10/n$$ $$1 < 1 +3/24n+1/24n^3 < 1+ 3/24n + 3/24n = 1+4/24n = 1+1/6n$$

So make sure that $1-10/n > \sqrt{1-\epsilon}$ and $1+1/6n < 1/\sqrt{1-\epsilon}$ and their ratio will be greater than $1-\epsilon$ but no larger than $1$.

Of course with better tools you will rewrite it as:

$${n^3(1-5/n + 13/n^2 - 5/n^3)\over 24n^3(1+3/24n+1/24n^3)} = {1\over24}{1-5/n + 13/n^2 - 5/n^3\over1+3/24n+1/24n^3}$$

where the second nominator and denominator $\to1$ as $n\to\infty$ and therefore the second fraction $\to1$ and the whole expression $\to1/24$ as $n\to\infty$.

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    You dont need the terms $n^3$, it is enough to see that $$\frac{n^3-5n^2+13n-5}{24n^3+3n+1}=\frac{1-5/n+13/n^2-5/n^3}{24+3/n^2+1/n^3}\to \frac1{24}$$ – 2017-01-03
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    @Masacroso Yes, I see that the limit is that immediately, but that won't help someone who doesn't. Besides it looked like he wanted a $\omega$-$\epsilon$ approach... – 2017-01-03