Question

Limits !

Answer 1

\[\lim_{n \rightarrow \infty}(\frac{ \pi }{ 2 }-\arctan(n^2))\ln(n!)\]

Answer 2

@ganeshie8

Answer 3

By wolfram test the limit equals 0
http://www.wolframalpha.com/input/?i=%5Clim_%7Bn+%5Crightarrow+%5Cinfty%7D+%28%28pi%2F2-%5Carctan%28n%5E2%29%29*%5Cln%28n%21%29%29

lets see how to work this out

Answer 4

wolfram test, lol

Answer 5

that test was invented by @freckles yesterday :D

Answer 6

\(\large\tt \begin{align} \color{black}{\lim_{n \rightarrow \infty} (\dfrac{\pi}{2}-\arctan(n^2))*\ln(n!))\\~\\
=\lim_{n \rightarrow \infty} (\dfrac{\pi}{2}-\arctan(n^2))*\lim_{n \rightarrow \infty} \ln(n!))\\~\\
=\left(\lim_{n \rightarrow \infty} (\dfrac{\pi}{2})-\lim_{n \rightarrow \infty} (\arctan(n^2))\right)*\lim_{n \rightarrow \infty} \ln(n!))\\~\\
=\left((\dfrac{\pi}{2})- (\arctan(\infty^2))\right)*\lim_{n \rightarrow \infty} \ln(n!))\\~\\
=\left((\dfrac{\pi}{2})- (\dfrac{\pi}{2})\right)*\lim_{n \rightarrow \infty} \ln(n!))\\~\\
=\left(0\right)*\lim_{n \rightarrow \infty} \ln(n!))\\~\\
=0}\end{align}\)

Answer 7

@mathmath333 - I thought you could only use the rule:\[\lim_{x\rightarrow a}f(x).g(x)=(\lim_{x\rightarrow a}f(x)).(\lim_{x\rightarrow a}g(x))\]if and only if both these individual limits exist: \(\lim_{x\rightarrow a}f(x)\), \(\lim_{x\rightarrow a}g(x)\)

Otherwise you would conclude that:\[\lim_{n\rightarrow\infty}\frac{1}{n}.n=(\lim_{n\rightarrow\infty}\frac{1}{n}).(\lim_{n\rightarrow\infty}n)=0.(\lim_{n\rightarrow\infty}n)=0\]

Answer 8

@asnaseer is right there's got to be other solution

Answer 9

we can use it under one roof

Answer 10

One way I can think of is:\[L=\lim_{n\rightarrow\infty}(\frac{\pi}{2}-\arctan(n^2)).\ln(n!)\]then we can say:\[e^L=e^{\lim_{n\rightarrow\infty}(\frac{\pi}{2}-\arctan(n^2)).\ln(n!)}=\lim_{n\rightarrow\infty}e^{(\frac{\pi}{2}-\arctan(n^2)).\ln(n!)}\]\[=\lim_{n\rightarrow\infty}(e^{(\frac{\pi}{2}-\arctan(n^2))})^{\ln(n!)}=1^\infty=1\]\[\therefore L=0\]But I am not sure if that is a valid approach - what do others think?

Answer 11

1^infinity is undetermination case @asnaseer

Answer 12

That is the /feeling/ I had as well which is why I was not sure of it :)

Answer 13

can \(ln (n!)\) be written as series

Answer 14

it can but not a very nice one unfortunately :(

Answer 15

we could swap the order of multiplication to get:\[e^L=\lim_{n\rightarrow\infty}e^{\ln(n!).(\frac{\pi}{2}-\arctan(n^2))}=\lim_{n\rightarrow\infty}(e^{\ln(n!)})^{(\frac{\pi}{2}-\arctan(n^2))}\]\[=\lim_{n\rightarrow\infty}(n!)^{(\frac{\pi}{2}-\arctan(n^2))}=\infty^0=1\]but that also doesn't feel right :(

Answer 16

http://www.wolframalpha.com/input/?i=series+of+ln+%28n%21%29+starting+at+n%3D%5Cinfty

Answer 17

this diverges so its \(\infty\)

Answer 18

series and this limit are not related

Answer 19

yes i was thinking of substituting the series for that dreaded lon n!

Answer 20

If it is easy try proving that the corresponding "series" converges by ratio or comparison tests and use below fact

series converges \(\implies\) sequence converges to 0

Answer 21

This is what I get for the series for \(\ln(n!)\) using wolframalpha: http://www.wolframalpha.com/input/?i=taylor+series+for+log%28n%21%29+at+n%3Dinfinity

Answer 22

really nobody ?

Answer 23

@eliassaab @Kainui

Answer 24

You can say \[\Large n! < n^n \\ \Large \ln(n!) < n \ln n\]