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Here are some other problems concerning converging of sequences:

Suppose $p(x) = 2x^{4}+x^{2}+3x+1$ and $q(x) = 3x^{4}+x^{3}+2x+3$. Define $a_n = \frac{p(n)}{q(n)}$ for $n \in \mathbb{Z}^{+}$. Prove that $a_n \to \frac{2}{3}$.

In other words, I want to show that $(a_{n}-\frac{2}{3})$ is a null sequence. Now we know that $a_n = \frac{ 2n^{4}+n^{2}+3n+1}{3n^{4}+n^{3}+2n+3}, \ n \in \mathbb{Z}^{+}$

So $a_n \leq \frac{2n^4+3n^2}{3n^4} = \frac{2}{3} + \frac{3}{n^2}$

Thus $a_n \to \frac{2}{3}$. Is this the best way of doing the problem (i.e. it is bounded and monotonic increasing)?

Let $[a_n, b_n]$ be a nested sequence of closed intervals such that $b_n-a_n \downarrow 0$ and let $(x_n)$ be a sequence such that $x_n \in [a_n, b_n]$ for all $n$. Prove that $(x_n)$ is convergent.

We know that $\bigcap (b_n-a_n) = 0$ for all $n$. So I think $\text{lim inf} \ \left( [a_n, b_n] \right) = \text{lim sup} \ \left([a_n, b_n] \right)$

This is how I think of it: $(x_n)$ becomes "trapped" in smaller and smaller intervals until it is forced to converge to a point. Is this the right way to think about it?

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    @Damien: The approach described by Beni Bogosel is easier, and quickly gets to the heart of the matter. If there were $-$ signs scattered in the polynomials, and large coefficients, estimates of the type you made could become unpleasant.2011-06-23

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For limits of the kind $\lim_{x \to \infty}\frac{f(x)}{g(x)}$ ($f,g$ polynomials) there is an algorithm. You need to know that $\lim_{x \to \infty}\frac{1}{x^k}=0,\ k >0$, so force out the greatest power of $x$ you can get from $f,g$.

Example in your case:

$\frac{x^4(2+\frac{1}{x^2}+\frac{3}{x^3}+\frac{1}{x^4})}{x^4(3+\frac{1}{x}+\frac{2}{x^3}+\frac{3}{x^4})}$

After simplifying the expressions you get some simple limits. If the degrees of polynomials are not the same, the result is $0$ if the greatest degree is in the denominator, or $\infty$ if it is the other way around.

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Concerning the second question, as Arturo pointed out in a comment above, $\lim _{n \to \infty } a_n = \lim _{n \to \infty } b_n $. Denote the common limit by $l$. Since $a_n \leq x_n \leq b_n$ for all $n$, by the squeeze theorem $\lim _{n \to \infty } x_n = l$.

Also, with regard to amWhy's comment above, it is worth recalling the Nested Interval Theorem (though you don't need it for the present question).

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Regarding the limit: Please see this FAQ thread: Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q$ and $P$ are polynomials.

It shows how to find $\lim_{n\rightarrow\infty}\frac{Q(n)}{P(n)}$ for any polynomials $P,Q$. (Which is summarized in Beni Bogosel's answer)