Problem Proving the Limit of $x_n = \frac{n+(-1)^n}{n+2}$

Question

I'm trying to prove the following limit using the algebra of limits and first principles, but I run into a problem.

First I find the limit using algebra of limits.

$$x_n = \frac{n+(-1)^n}{n+2} = \frac{1+\frac{(-1)^n}{n}}{1 + \frac{2}{n}} \implies \lim_{n \rightarrow \infty} x_n = 1$$

Next using first principles. Fix $\epsilon > 0$, $N \in \mathbb{N}$

$$n\geq N \implies \left|\frac{n+(-1)^n}{n+2} - 1\right| < \epsilon$$

However when I try set a $n$ to remove the absolute value there doesn't appear to be $n$ (except for something nearing $\infty$) that ensures

$$\left|\frac{n+(-1)^n}{n+2} - 1\right| =\frac{n+(-1)^n}{n+2} - 1 $$

Any help would be much appreciated!

Answer 1

$|{n+(-1)^n \over n+2} -1| = |{2-(-1)^n \over n+2}|\le {3 \over n}$.

Answer 2

You have that $$ x_n = \frac{n} {{n + 2}} + \left( { - 1} \right)^n \frac{1} {{n + 2}} $$ Since $b_n=(-1)^n$ is a bounded sequence and since $$ \mathop {\lim }\limits_{n \to + \infty } \frac{1} {{n + 2}} = 0 $$ you have that $$ \mathop {\lim }\limits_{n \to + \infty } \left( { - 1} \right)^n \frac{1} {{n + 2}} = 0 $$ Since $$ \mathop {\lim }\limits_{n \to + \infty } \frac{n} {{n + 2}} = 1 $$ you have that $$ \mathop {\lim }\limits_{n \to + \infty } x_n = \mathop {\lim }\limits_{n \to + \infty } a_n + \mathop {\lim }\limits_{n \to + \infty } b_n = 1 + 0 = 1 $$

Answer 3

Note that $n-1<n+1<n+2$, therefore $$\frac{n-1}{n+2}<\frac{n+1}{n+2}<1.$$ That is $$\left|\frac{n+(-1)^n}{n+2}-1\right|=1-\frac{n+(-1)^n}{n+2}.$$