August 18th, 2019, 09:11 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91  Hard problem
Evaluate $\displaystyle L(s)=\lim_{n\rightarrow \infty } s^{n} \: $ , $\displaystyle s\in \mathbb{R}^{+} $.

August 18th, 2019, 11:08 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,272 Thanks: 942 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \lim_{n \to \infty} s^n \to \begin{cases} \infty & 1 < s \\ 1 & 1 = s \\ 0 & 1 > s \end{cases} $ Dan  
August 19th, 2019, 06:54 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 
Check this one : $\displaystyle L(s)=s\cdot \lfloor s \rfloor $. L=s[s] . But it is not working for s>1 . Last edited by idontknow; August 19th, 2019 at 07:25 AM. 
August 19th, 2019, 08:13 AM  #4  
Senior Member Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91  Quote:
$\displaystyle L(s)=s\lfloor s \rfloor +\frac{s\lfloor s \rfloor 1}{\lfloor s^{1} \rfloor } $$\displaystyle \lfloor s \rfloor .$ Last edited by idontknow; August 19th, 2019 at 08:24 AM.  
August 19th, 2019, 09:14 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,972 Thanks: 2222 
On that assumption, $\displaystyle L(s) = \frac{\lfloor s \rfloor}{\lfloor s^{1} \rfloor}$ is simpler.


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