# [SOLVED]Limits of infinity

#### dwsmith

##### Well-known member
$\lim\limits_{n\to\infty}\frac{3n+5}{2n+7}=\frac{3}{2}$

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

#### Opalg

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

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
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
$\lim\limits_{n\to\infty}\frac{3n+5}{2n+7}=\frac{3}{2}$

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
Whenever $n > N$
$$\frac{2}{2n + 7} < \epsilon$$

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

#### dwsmith

##### Well-known member
That's not what I got for $$\left|\frac{3n+5}{2n+ 7}- \frac{3}{2}\right|$$. How did you get that?