Proving $\lim_{n\to\infty} n^2/(n^2 + n) = 1$ using $\epsilon-N$.

Question

I would like someone to verify my $\epsilon-N$ proof that $\lim_{n\to\infty} \dfrac{n^2}{n^2 + n}$ = 1.

Proof:

Let $\epsilon > 0 $, and choose $N = \lceil1/\epsilon\rceil + 1$. Then for all $n > N$, we have

$$\left|\frac{n^2}{n^2 + n} - 1 \right| = \left|\frac{-n}{n^2 + n}\right| = \frac{n}{n^2 + n} \leq \frac{n}{n^2} = 1/n.$$

But since we have $n > N = \lceil1/\epsilon\rceil + 1 > \lceil1/\epsilon\rceil \geq 1/\epsilon$, it follows that $\epsilon > 1/n$. Thus, $1/n < \epsilon$ as we desired to show. Therefore, we conclude that the limit equals $1$.

Answer 1

Yes, it's correct. But another approach could be:

Let $\epsilon>0$. For the archimedean property there is $N\in \mathbb{N}$ such as, $$\frac{1}{N}<\epsilon.$$ Let $n\geq N$, then, $$\left|\dfrac{n^2}{n^2+n}-1\right|=\dfrac{n}{n^2+n}\leq \dfrac{1}{n}\leq \dfrac{1}{N}<\epsilon$$ So, $$\lim_{n\to \infty}\dfrac{n}{n^2+n}=1.$$

Answer 2

Note that $$ \frac{n}{n^2+n}=\frac{1}{n+1} $$ Since $$ \frac{1}{n+1}<\varepsilon $$ if and only if $$ n>\frac{1}{\varepsilon}-1 $$ you can just take $N=\lceil \varepsilon^{-1}-1\rceil$ (least integer greater than or equal to $\varepsilon^{-1}-1$), which exists by the Archimedean property.

Of course, any integer $N>1/\varepsilon$ would do as well.