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I want to prove $\displaystyle \lim_{n \rightarrow \infty}n^3 - 4n^2 -100n = \infty$ using the definition $\forall c \exists N \in \mathbb{N} ,s.t.\ a_n>c\ \forall n\geq N$.

I'm not having the best luck in thinking about this one. I think that I cannot use the limit sum properties if the sequence goes to infinity. So I can't look at the parts. I know that like convergence I assume that I'm given $c$ and show that I can find an N that makes the inequality hold. But the function $n^3 - 4n^2 -100n$ has a sign change around n = 12. I'm not sure how that affects things if it does at all.

Anyways. If I get $c$ from someone and look at $N > (c+13)$. Then I say $n^3 - 4n^2 -100n > N^3 - 4N^2 -100N = (c+13)^3 - 4(c+13)^2 -100(c+13)$ and then $(c+13)^3 - 4(c+13)^2 -100(c+13)=c^3 + 35 c^2+303 c+ 221 > c$ Does this work or have I screwed up somewhere?

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Archemedian property – leo Feb 17 '12 at 4:59
Observe that $n^3 = n\cdot n^2$. Now assume that $n > 104$, that is, write $n = 104 + k$, where $k\geq 1$. Then we get: $n^3 = (104 + k)n^2 = 104n^2 + k(104 + k)^2$. Thus we obtain: $n^3 - 4n^2 - 100n = 104n^2 + k(104 +k)^2 + 4n^2 - 100n > k(104 +k)^2$. What can you conclude from that? – William Feb 17 '12 at 5:04
up vote 7 down vote accepted

$n^3 - 4n^2 - 100n = n(n^2-4n-100)$ and this is greater than $n$ if $n^2-4n-100 > 1$.

Now, $n^2 - 4n - 100 > 1$ iff $n(n-4) >101$ and some trial and error shows that $n>13$ will satisfy the previous inequality.

Soo... For every $n \geq 13$ $a_n > n \dots$

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The calculation in your last paragraph really doesn’t make much sense. It’s true that $$(c+13)^3 - 4(c+13)^2 -100(c+13)=c^3 + 35 c^2+303 c+ 221\;,$$ but it isn’t necessarily greater than $c$, and it doesn’t tell you anything about $$N^3-4N^2-100N$$ for $N>c+13$. Here’s a slightly more concrete version of the approach suggested by Daniel Pietrobon.

Consider the function $f(x)=x^3-4x^2-100x=x(x^2-4x-100)$; it will be convenient to let $g(x)=x^2-4x-100=(x-2)^2-104$. Clearly $g(x)$ is increasing for $x>2$. As you’ve already noticed, $f(12)<0$ and $f(13)>0$, so $g(13)>0$. In fact, $g(13)=17$, so $g(x)\ge 17$ for all $x\ge 13$, and therefore $xg(x)\ge 17x$ for all $x\ge 13$. (We can multiply the inequality $g(x)\ge 17$ by $x$, since we’re looking only at $x\ge 13$, and they are certainly positive.) In other words, $$f(x)\ge 17x\text{ for all }x\ge 13\;,$$ and now it should be very easy to complete the argument.

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A different way to calculate the limit, which generalizes easily to any polynomial, is to factor out the leading term first: $$n^3-4n^2-100n=n^3\Bigl(1-{4\over n} -{100\over n^2} \Bigr).$$ Pick $m$ sufficiently large so that $1-{4\over m} -{100\over m^2}\ge {1\over 2}$ (of course, here, I'm assuming that $\lim\limits_{n\rightarrow\infty}{a\over n^\alpha}=0$ for $\alpha>0$ is already in your tool bag). Then, for $n\ge m$ $$n^3-4n^2-100n\ge{ n^3\over 2} .$$ And now you need only select $N=\max\{ \lceil\,\root 3\of{2c}\,\rceil ,m\}$ to achieve your end.

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You undoubtedly know intuitively what's happening: After a while, $n^3$ is so big that taking away $4n^2$ and $100n$ still leaves $n^3-4n^2-100n$ big. The solution below makes this intuition precise.

Let's make sure that $4n^2<(1/4)n^3$ and $100n<(1/4)n^3$. Then when we take them away from $n^3$, it will still leave plenty.

The first inequality is true past $16$. The second inequality is true past $20$. So past $20$ we have $$n^3-4n^2-100n<n^3-\frac{n^3}{4}-\frac{n^3}{4}=\frac{n^3}{2}.$$

So as long as we also make $n>20$, we can be sure that $a_n>c$ if $n>\sqrt[3]{2c}$. We can, for example, take $N=\max\left(20, \lfloor\sqrt[3]{2c} \rfloor\right)+1$.

The estimate we made above brings out the long run $n^3$-like behaviour of $a_n$.

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