Formal definition of limit of a sequence

Question

The formal definition of limit of a sequence says that for every $\epsilon>0$, there is a number $M>0$ such that if $n > M$ it is true that $|(a_n) - L| < ε$

In this particular case I have to demonstrate by the definition that this limit exist: $$\lim \limits_{n \to \infty}\frac{(n^2+4)}{(n-1)(n+2)(n-3)}=0$$
I have made the following reasoning: I have a hypothesis that is "for all $ n> M $" and a thesis that says "it is true that $ | (a_n) - L | <ε$" but contrary to initiate by the hypothesis to arrive to the thesis, I can initiate by the thesis to arrive at the hypothesis if the steps are reversible.

So, the firts step would be: $$\left\lvert\frac{(n^2+4)}{(n-1)(n+2)(n-3)}\right\rvert <\epsilon $$
And try to take out of the absolute value the expression inside it, then isolate 'n' in terms of ε. We call it equation (1)

The second step would be: find a $M>0$, for that reason conveniently we say that $M>$ equation (1)

The third step would be: By the Archimedean property we say that there is a $n>$$M>$ equation (1) and then, by the transitivity of the order relation we say that $n>$ equation (1)

The fourth step would be: Isolate 'ε' [that is in equation (1)] in terms of 'n'; call it equation (2)

Final step would be: Take in the absolute value in equation (2) and notice that gives us back the next form: $$\left\lvert\frac{(n^2+4)}{(n-1)(n+2)(n-3)}\right\rvert <\epsilon $$

The problem is that is difficult to do the first step because, is hard to take out of the absolute value the expression inside it. And then isolate 'n' in terms of ε (even for wolfram alpha). I am asking for any help or a solution for this exercise. Thanks in advance

Answer 1

For $n\ge 4$ we have

$$\frac{n^2+4}{(n-1)(n+2)(n-3)}<\dfrac{n^2+4}{(n+2)^3}<\dfrac{2n^2}{(n+2)^3}<\dfrac{2(n+2)^2}{(n+2)^3}=\dfrac{2}{n+2}.$$ Thus $$\epsilon=\dfrac{2}{n+2}$$ works.

Answer 2

$\lim_{n \rightarrow \infty}a_n=l$ if $\forall \epsilon >0,\exists N \in \mathbb{N}$ such that $|a_n-l| \leq \epsilon, \forall n \geq N$

Lets denote your sequence $a_n$

Let $\epsilon >0$.

For $n \geq 4$, by doing some calculations we see that $|a_n| \leqslant \frac{1}{n}$

From the Archimedian propeprty exists $N \in \mathbb{N}$ such that $$\frac{1}{N} < \epsilon \Rightarrow N> \frac{1}{\epsilon}$$

Take $N=[\frac{1}{\epsilon}]+1$ where $[.]$ is the integer part of a number and for which we have that $\forall x$ $$[x] \leq x \leq [x]+1$$ $$x-1 \leq [x] \leq x $$

Now notice that this $N=[\frac{1}{\epsilon}]+1$ is the minimum natural number with the property that whenever $$n \geq N \Rightarrow \frac{1}{n}< \epsilon $$

Also form the definition of limit this $N$ depends on $\epsilon$

Now can you see now how the finals step goes?

Also your sequence is a little complicated to find such $N$ that depends on $\epsilon$ so we can bound $a_n$ with as simpler sequence $b_n$ which will allow us to prove more easily the existence of such $N$.