Prove $x_n=\frac{n+(-1)^n}{n+2}$ has limit 1 from first principles

Question

Prove that that the sequence $x_n=\frac{n+(-1)^n}{n+2}$ has limit 1 from first principles. I need help finding a suitable $N(\epsilon)$.

My attempt:

Fix $\epsilon$>0. We want to find N$\in$$\mathbf{N}$ such that:

$n>N \implies \lvert\frac{n+(-1)^n}{n+2} - 1\rvert<\epsilon $

$ \implies\lvert\frac{n+(-1)^n}{n+2} - \frac{n+2}{n+2}\rvert<\epsilon$

$ \implies\lvert\frac{(-1)^n -2}{n+2}\rvert<\epsilon$

Now this is where I'm unsure, I don't know what to do with the $(-1)^n$. Do I want to make $\lvert\frac{3}{n+2}\rvert<\epsilon$ and then choose $N > \frac{3-2\epsilon}{\epsilon}$? Or is this an incorrect approach?

Answer 1

Yes, this is correct, especially when $n$ is odd, so that $$\left\lvert\frac{(-1)^n -2}{n+2}\right\rvert = \frac{3}{n+2}<\epsilon.\tag{*}\label{*}$$

Note that in the definition of limit of a sequence, it reads

$\forall\,\epsilon > 0, \exists N \in \Bbb{N}$ such that ${\bf \forall n>N}$, $|a_n - L| < \epsilon$,

so you need inequality \eqref{*} verified by all $n > N$, especially when $n$ is odd.

Answer 2

\begin{align*} \left|\dfrac{n+(-1)^{n}}{n+2}-1\right|&=\left|\dfrac{(-1)^{n}-2}{n+2}\right|\\ &\leq\dfrac{3}{n+2}\\ &<\dfrac{3}{N+2}\\ &<\dfrac{3}{N}\\ &<\epsilon \end{align*} for $n\geq N$, where $N$ is such that $3/N<\epsilon$ by Archimedean.