I for the most part have this completed but I have a small question and thus checking if I did this correctly.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

Given the Sequence = [tex]\frac{n}{n+2}[/tex] [tex]\approx_{\epsilon}[/tex] 1 , for n >> 1

Show what index if [tex]\epsilon[/tex] = .001

" " if [tex]\epsilon[/tex] = .000002

" " for any [tex]\epsilon[/tex] > 0

2. Relevant equations

N/A

3. The attempt at a solution

for [tex]\epsilon[/tex] = .001 I did:

[tex]\left|\frac{n}{n+2}-1\right|< .001[/tex]

[tex]\left|\frac{-2}{n+2}\right| < .001[/tex]

[tex]\frac{2}{n+2} < .001[/tex]

[tex]\frac{2}{n+2} < \frac{1}{1000}[/tex]

[tex]\frac{n+2}{2} > 1000[/tex]

[tex]n+2 > 2000[/tex]

[tex]n > 1998[/tex]

The only issue I have with this one and the next one is that they aren't decreasing and I'm not sure if it needs to be or not:

[tex]\frac{1999}{1999+2} = .9990004998[/tex]

[tex]\frac{2000}{2000+2} = .9990009990[/tex]

for [tex]\epsilon[/tex] = .000002 I did:

[tex]\left|\frac{n}{n+2}-1\right|< .000002[/tex]

[tex]\left|\frac{-2}{n+2}\right| < .000002[/tex]

[tex]\frac{2}{n+2} < .000002[/tex]

[tex]\frac{2}{n+2} < \frac{2}{1000000}[/tex]

[tex]\frac{n+2}{2} > \frac{1000000}{2}[/tex]

[tex]n+2 > 1000000[/tex]

[tex]n > 999998[tex]

for any [tex]\epsilon[/tex] > 0

[tex]\left|\frac{n}{n+2}-1\right|< \epsilon[/tex]

[tex]\left|\frac{-2}{n+2}\right| < \epsilon[/tex]

[tex]\frac{2}{n+2} < \epsilon[/tex]

[tex]\frac{n+2}{2} > \frac{1}{\epsilon}[/tex]

[tex]n+2 > \frac{2}{\epsilon}[/tex]

[tex]n > \frac{2}{\epsilon}-2[/tex]

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# Finding indexes.

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