limit ln(n/(2n+5)) as n -> infty

[thebenji]

how do i evaluate the limit as n goes to infinity of

ln[n/(2n+5)]

?

 

[galactus]

try rewriting:

\(\displaystyle \L\\ln(\frac{n}{2n+5})=ln\left(\frac{1}{2}-\frac{5}{2(2n+5)}\right)\)

\(\displaystyle \L\\\lim_{n\to\infty}[ln\left(\frac{1}{2}-\frac{5}{2(2n+5)}\right)]\)

Now, see what the limit is?.

 

[stapel]

I would try setting y equal to ln(x/(2x + 5)), and then looking at the limit of e^y as x goes to infinity.

Eliz.

 

[soroban]

Re: limit ln(n/(2n+5))

Hello, thebenji!

How do I evaluate: \(\displaystyle \,\L\lim_{n\to\infty} \ln\left(\frac{n}{2n\,+\,5_\right)\)

The same way you treat most rational functions . . .

Divide top and bottom by \(\displaystyle n:\)

. . \(\displaystyle \L\lim_{n\to\infty}\ln\left(\frac{\frac{n}{n}}{\frac{2n}{n}\,+\,\frac{5}{n}}\right) \;= \;\lim_{n\to\infty}\ln\left(\frac{1}{2\,+\,\frac{5}{n}\right)\)

. . \(\displaystyle \L=\;\ln\left(\frac{1}{2\,+\,0}\right)\;=\;\ln\left(\frac{1}{2}\right) \;=\;-\ln(2)\)