MHB What is the limit at infinity of (3n+5)/(2n+7)?

[Dustinsfl]

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

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

 

[Opalg]

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

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 $n > N$.

 

[Dustinsfl]

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]

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

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]

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?

 

[Dustinsfl]

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

I can't add or subtract. :)