On proving $\lim \frac{n^2}{n^2+n+1} = 1$

Question

Prove that $\lim \frac{n^2}{n^2+n+1} = 1$

Let $\varepsilon > 0$ and let $N = \frac{1}{\varepsilon}.$ Then $n > N$ implies $n > \frac{1}{\varepsilon} \implies \frac{1}{n} < \varepsilon.$

But $\displaystyle \frac{1}{n} = \frac{n+1}{n(n+1)} = \frac{n+1}{n^2+n} > \frac{n+1}{n^2+n+1} = \bigg|\frac{n^2}{n^2+n+1}-1\bigg|.$

Therefore $\bigg|\frac{n^2}{n^2+n+1}-1\bigg| < \varepsilon$, hence $\displaystyle \lim \frac{n^2}{n^2+n+1} = 1.$

Could someone please verify whether the above is correct.

Answer 1

Your stream of thought is right; just notice that $1/\varepsilon$ is not necessarily an integer, which matters if we are talking about a sequence, which is a map defined on a subset of the set of all integers. Instead, you may write $N := \lceil 1/\varepsilon \rceil + 1$, by which you can ensure the choice of $N$ to be an integer. However, this depends on how one defines convergence of sequence; and it is really not a big deal as long as you are consistent in what you are talking about.

Answer 2

if $\forall n.a_n \gt 0$ $$ \lim_{n\to\infty} a_n = 1 \Leftrightarrow \lim_{n\to\infty} \frac1{a_n} = 1 $$ set $$ a_n = \frac{n^2+n+1}{n^2} = 1+\frac1{n}+\frac1{n^2} $$ so for $n \gt 1$ $$ 0 \lt a_n-1 \lt \frac2n $$