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- Thread starter dwsmith
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- Feb 7, 2012

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Instead of a delta epsilon proof, you need to use an N epsilon proof. In other words, given $\varepsilon>0$, you need to find $N$ such that $\Bigl|\frac{3n+5}{2n+7}-\frac{3}{2}\Bigr| < \varepsilon$ whenever $n > N$.$\lim\limits_{n\to\infty}\frac{3n+5}{2n+7}=\frac{3}{2}$

How does one use a delta epsilon proof for a limit at infinity?

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Whenever $n > N$Instead of a delta epsilon proof, you need to use an N epsilon proof. In other words, given $\varepsilon>0$, you need to find $N$ such that $\Bigl|\frac{3n+5}{2n+7}-\frac{3}{2}\Bigr| < \varepsilon$ whenever $n > N$.

$$

\frac{2}{2n + 7} < \epsilon

$$

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- Feb 13, 2012

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The easiest way is to set $\displaystyle x=\frac{1}{n}$ and to apply the delta epsilon proof to...$\lim\limits_{n\to\infty}\frac{3n+5}{2n+7}=\frac{3}{2}$

How does one use a delta epsilon proof for a limit at infinity?

$\displaystyle \lim_{x \rightarrow 0} \frac{\frac{3}{x}+5}{\frac{2}{x}+7}$

Kind regards

$\chi$ $\sigma$

- Jan 29, 2012

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That's not what I got for [tex]\left|\frac{3n+5}{2n+ 7}- \frac{3}{2}\right|[/tex]. How did you get that?Whenever $n > N$

$$

\frac{2}{2n + 7} < \epsilon

$$

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I can't add or subtract.That's not what I got for [tex]\left|\frac{3n+5}{2n+ 7}- \frac{3}{2}\right|[/tex]. How did you get that?