<s>[INST] Show that $\lim_{n\to\infty}\frac{n+1}{n-2}=1$ 
By epsilon-delta definition, show that $\displaystyle\lim_{n\to\infty}\frac{n+1}{n-2}=1$.

I start with $\left|\frac{n+1}{n-2}-1\right|=\left|\frac 3 {n-2}\right|$, but I don't know how to carry on. Thank you.
 [/INST] Write $\frac{n+1}{n-2} = 1+\frac{3}{n-2}$. Then $\left|\frac{n+1}{n-2} -1\right| = \left|\frac{3}{n-2}\right|$. Let $\epsilon>0$ and choose $N > 2+\frac{3}{\epsilon}$. Then if $n \ge N$, you have $\left|\frac{3}{n-2}\right|< \epsilon$.
To get a lower bound for $N$: We want $\left|\frac{3}{n-2}\right|< \epsilon$. This will be true if $3 < |n-2| \epsilon$, or equivalently $\frac{3}{\epsilon} < |n-2|$. Since we are interested in $n \to \infty$, we ask that $n$ satisfy $\frac{3}{\epsilon} < n-2$, or equivalently, $2+\frac{3}{\epsilon} < n$. So, if we pick $N > 2+\frac{3}{\epsilon}$, then if $n \ge N$, we will have the desired result.
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