My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree3Thanks
  • 2 Post By topsquark
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
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}^{+} $.
idontknow is offline  
 
August 18th, 2019, 11:08 AM   #2
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,272
Thanks: 942

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by idontknow View Post
Evaluate $\displaystyle L(s)=\lim_{n\rightarrow \infty } s^{n} \: $ , $\displaystyle s\in \mathbb{R}^{+} $.
This is pretty much straightforward so I'm not going to supply a proof, just the answer.
$\displaystyle \lim_{n \to \infty} s^n \to \begin{cases} \infty & 1 < s \\ 1 & 1 = s \\ 0 & 1 > s \end{cases} $

-Dan
Thanks from romsek and idontknow
topsquark is offline  
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.
idontknow is offline  
August 19th, 2019, 08:13 AM   #4
Senior Member
 
Joined: Dec 2015
From: somewhere

Posts: 642
Thanks: 91

Quote:
Originally Posted by idontknow View Post
Check this one : $\displaystyle L(s)=s\cdot \lfloor s \rfloor $.
L=s[s] . But it is not working for s>1 .
Assuming $\displaystyle 1/0=\infty $ , the equality holds true .
$\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.
idontknow is offline  
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.
Thanks from idontknow
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
hard, problem



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Hard problem Ionika Algebra 0 February 5th, 2014 11:18 AM
Hard problem Cephal Calculus 2 December 27th, 2013 05:18 PM
A really hard problem! math221 Calculus 8 March 17th, 2013 05:51 PM
hard problem?? john mraz Advanced Statistics 0 June 15th, 2010 12:32 AM
hard problem nguoidep_68 Algebra 3 October 31st, 2007 03:11 PM





Copyright © 2019 My Math Forum. All rights reserved.