So, I'm stuck on this question, and have been working on it for a few hours, probably because I'm not 100% in understanding the definition. But here's the question:
Using the $\epsilon$ – N definition of the limit prove that,
$\lim_{x \to ∞}{1\over(n^2+n)}$ = $0$ (sorry I don't know how to write fractions in limits).
In other words, given $\epsilon$ > $0$, find explicitly a natural number N which satisfies the statement in the definition of the limit.
So the definition I received in lectures is:
Let $(a_n)_1^∞$ ($1$ should be $n=1$, I couldn't get it to work, sorry) be a sequence of real numbers $(a_n)_1^∞$ = {$a_1, a_2, a_3,…$} .
Definition: We write $\lim_{n \to ∞} a_n = a$, and say the limit of $(a_n)_1^∞$ equals $a$, if for every $\epsilon$ > 0 there exists N ∈ $\mathbb{N}$ such that if $n ≥ N$, then $|a_n – a| < \epsilon$.
So far this is what I've done:
|$\frac{1}{n^2+n}$ + 1| < $\epsilon$
$\therefore$ |$\frac{1-n^2+n}{n^2+n}$ + 1| < $\epsilon$
I did try other ways like disregarding the 'n' in the denominator since $n^2$ is more significant then it, but I don't think I'm supposed to do that. I'm really just stuck on what to do next as I don't really understand the process. Any help would be GREATLY appreciated, thanks! 🙂
Best Answer
Let us use the definition you stated in your post. (BTW, the attempt you made is wrong).
Let $\epsilon>0$. By using the Archimedean Property, we can find $N\in\Bbb N$ such that $\frac{1}{N}<\epsilon$. Thus, if $n\geq N$ then we get $$\bigg|\frac{1}{n^2+n}-0\bigg|=\frac{1}{n^2+n}<\frac{1}{n}\leq \frac{1}{N}<\epsilon.$$ Hence, $$\lim_{n\to\infty}\frac{1}{n^2+n}=0.$$