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[SOLVED] Limits of infinity

dwsmith

Well-known member
Feb 1, 2012
1,673
$\lim\limits_{n\to\infty}\frac{3n+5}{2n+7}=\frac{3}{2}$

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

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
$\lim\limits_{n\to\infty}\frac{3n+5}{2n+7}=\frac{3}{2}$

How does one use a delta epsilon proof for a limit at infinity?
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 $n > N$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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 $n > N$.
Whenever $n > N$
$$
\frac{2}{2n + 7} < \epsilon
$$

Like this?
 

chisigma

Well-known member
Feb 13, 2012
1,704
$\lim\limits_{n\to\infty}\frac{3n+5}{2n+7}=\frac{3}{2}$

How does one use a delta epsilon proof for a limit at infinity?
The easiest way is to set $\displaystyle x=\frac{1}{n}$ and to apply the delta epsilon proof to...

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


Kind regards


$\chi$ $\sigma$
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Whenever $n > N$
$$
\frac{2}{2n + 7} < \epsilon
$$

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

dwsmith

Well-known member
Feb 1, 2012
1,673
That's not what I got for [tex]\left|\frac{3n+5}{2n+ 7}- \frac{3}{2}\right|[/tex]. How did you get that?
I can't add or subtract. :)