Question

Help?

Answer 1

http://i.imgur.com/nACXrOe.png

Answer 2

are you allowed to use l'hopital's rule (i hope)?

Answer 3

\[\lim_{n \rightarrow 00} e^{nln(\frac{(n+3)}{(n+1)}}\]

Answer 4

if so, then good
if not then write it as
\[1+\frac{2}{n+1}\]

Answer 5

then you get pretty much instantly that
\[\lim_{n\to \infty}\left(1+\frac{2}{n+1}\right)^n=e^2\]

Answer 6

Yes, i can use l'hoptal's rule BUT i'm stuck with the basic logs, lol

Answer 7

ok here is the gimmick

Answer 8

you need
\[\lim_{n\to \infty}n\ln(\frac{n+3}{n+1})\] which looks like
\[\infty\times 0\]

Answer 9

Yes

Answer 10

I wish I can medal you again, @satellite73

Answer 11

the trick is to make it look like
\[\frac{0}{0}\] by rewriting it as \[\frac{\ln(\frac{n+3}{n+1})}{\frac{1}{n}}\]

Answer 12

THEN use l'hopital's rule

Answer 13

OH

Answer 14

Omg, thanks satelite

Answer 15

I took it too far

Answer 16

bunch of algebra involved once you take the derivative , but eventually you will get that the limit is 2, making your original limit \(e^2\)

Answer 17

Yep ^_^

Answer 18

you can see the \(e^2\) more clearly here \[\lim_{n\to \infty}\left(1+\frac{2}{n+1}\right)^n=e^2\]

Answer 19

since \[\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n\] is one definition of \(e^x\)

Answer 20

good luck, and your welcome, no problem
one good question in before i retire for the night

Answer 21

@nincompoop thank you as well!

Answer 22

Yeah, i forgot the limit definition of e^x
I need to write this down. Last time i worked with limits was last semester..
Thanks a lot @satellite73, my brain went back to its normal stance

Answer 23

yw

Answer 24

good job, @Zale101

Answer 25

Thanks, lol if i keep making these simple mistakes i will fail the series chapter.