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.
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Brian M. Scott
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JSCB
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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.
copper.hat
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But we don't know if n-2>0 ... – JSCB Apr 11 '13 at 07:23
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You do. Since $n \ge N$, you have $n-2 \ge \frac{3}{\epsilon} > 0$. – copper.hat Apr 11 '13 at 07:25
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I get it, thanks – JSCB Apr 11 '13 at 07:26
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How do you think to choose that N? – JSCB Apr 11 '13 at 07:34
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I just worked backwards from the requirement $\frac{3}{n-2} < \epsilon$. – copper.hat Apr 11 '13 at 07:36
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Can you show me the whole process of how you work backward please? – JSCB Apr 11 '13 at 07:41
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I added a paragraph to the end above... – copper.hat Apr 11 '13 at 07:53
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Actually don't we have to consider all kind of value of n? Will there be problem if we just focus on n-2>0 ? – JSCB Apr 11 '13 at 07:58
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No. You just need to show that for any $\epsilon>0$ that there exists some $N$ such that the above condition holds for $n \ge N$. – copper.hat Apr 11 '13 at 08:01
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You are very welcome! – copper.hat Apr 11 '13 at 08:26
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$\left|\frac{3}{n-2}\right|<\epsilon\Leftrightarrow n-2>\frac{3}{\epsilon}$, so you just need to choose $n>\left\lceil{2+\frac{3}{\epsilon}}\right\rceil$.
Brian M. Scott
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