Let $\lbrace a_n \rbrace$ be a real sequence. We say $\lim_{n\to\infty} a_n=\infty$ provided that:

$$\forall K>0, \exists N\in \mathbb{N} \forall n \ge N:a_n>K$$

Use this definition, as well as the Archimedean property, to prove that: $$\lim_{n\to\infty}n^3-4n^2-99n=\infty$$

In this context, the Archimedean property is defined as:

$$\forall c\in\mathbb{R} \exists n\in\mathbb{N}:n>c$$

I'm a little lost on how to use this particular definition of the limit. Perhaps if the roots were easier, it wouldn't be so difficult, but I'm not making progress even when I try simpler versions. So how could you prove this using that definition in a fairly simple way?

  • $\begingroup$ I don't understand why you consider this definition of limit "unusual". It is the usual one, according to me. $\endgroup$ – Giuseppe Negro Feb 26 '18 at 12:00

Write : $a_n = n^3 - 4n^2 - 99n = n(n^2 - 4n - 99)$.

Now, use a trivial bound : $n \geq 1$ for all $n \in \mathbb N$.

This gives $a_n \geq n^2 - 4n - 99 = (n-2)^2 - 103$.

Therefore, $a_n \geq (n-2)^2 - 103$ for all $n$. We will call this $(*)$

A small observation : if $a, b > 2$ are two numbers and $a > b$, then $a-2 >b-2$, so $(a-2)^2 > (b-2)^2$.

Now, we will start from the definition of the limit. So, let $K$ be any real number.

By the Archimedean property applied to $K$, we get there is some natural number $N_0$ such that $N_0 > \sqrt{K + 103} + 2$, since $\sqrt{K+103} + 2$ is also just a real number. Let $N = \max(N_0,2)$. Then $N \geq N_0$, so the same inequality also applies with left hand side as $N$.

Rearrange the above to get $(N-2)^2 - 103 > K$.

Now, take any $n > N$. We have the following inequalities: $$ a_n \geq (n - 2)^2 - 103 \geq (N-2)^2 - 103 > K $$ Where we used $(*)$ for the first inequality, and the small observation for the second inequality.

Since $K$ was any large enough real number, we have by the definition of limit that $\lim a_n = \infty$.


In general, using the same technique, you can show that for any polynomial in a single variable, it is enough to look at the degree of the polynomial, along with the sign of the coefficient of the highest degree monomial, to conclude the limit of the polynomial as the variable goes to infinity, plus or minus. In particular, if the degree is non-zero, as it is in our case, then you can conclude that the expression goes to infinity, where infinity has the same sign as the coefficient of the highest degree monomial. Here, the coefficient is one, which is positive. A similar logic extends to fractions with numerator and denominator being polynomials : these are called rational functions. If you like I can let the cat out of the bag on these functions as well.

  • $\begingroup$ I don't quite understand why defining another variable $N$ as the maximum of $N_0$ and $2$ was necessary in your example. Is it not true that $N_0$ is always greater than 2? $\endgroup$ – jippyjoe4 Feb 24 '18 at 5:19
  • $\begingroup$ Well , it might be. I was trying to be safe. You may as well be correct. If I can conclude it, and I am pretty sure I will be able to, then we can edit the answer and remove the unnecessary condition from the answer. Oh, it is, so we can remove it. On that note, you can read the edit : it contains more useful information on problems similar to the one you were dealing with. $\endgroup$ – астон вілла олоф мэллбэрг Feb 24 '18 at 5:28

Note that $$ n^3-4n^2-99n=n(n^2-4n-99) $$ and that $n^2-4n-99=n(n-4)-99>1$ as soon as $n>12$. Thus, for $n>12$ you have $n^3-4n^2-99n>n$.


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