February 6th, 2012, 12:36 PM  #1 
Member Joined: Jan 2012 Posts: 63 Thanks: 0  continuous for 1/x
For the function f(x)=1/x, prove that with ? =1, there does not exist a ?>0 s.t. the definition for continuity "for all ? > 0, exist a ? > 0 s.t. xc<? then f(x)f(c)<?" holds for all points c and x in (0,infinity).

February 7th, 2012, 06:00 AM  #2  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,947 Thanks: 1139 Math Focus: Elementary mathematics and beyond  Re: continuouse for 1/x Quote:
 
February 9th, 2012, 05:43 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,753 Thanks: 2136 
No, infinity will do. If such ? > 0 exists, let c = ?/(2? + 2) and x = c + ?/2 = (?² + 2?)/(2? + 2), so that x  c = ?/2 < ?. f(x)  f(c) = (2? + 2)/(?² + 2?)  (2? + 2)/? = (2? + 2  (2? + 2)(? + 2) + (?² + 2?))/(?² + 2?)  1 = (2? + 2  2?²  6?  4 + ?² + 2?)/(?² + 2?)  1 = ( ?²  2?  2)/(?² + 2?)  1 = ((? + 1)² + 1)/(?² + 2?) + 1 > 1. Hence f(x)f(c) < ? does not hold when ? = 1, contradicting the original supposition. The question didn't giving the correct definition of continuity. 

Tags 
1 or x, continuous 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
continuous  frankpupu  Calculus  4  January 31st, 2012 05:17 AM 
Proving that a space is continuous, continuous at 0, and bdd  thedoctor818  Real Analysis  17  November 9th, 2010 08:19 AM 
continuous map  guroten  Real Analysis  2  November 5th, 2010 01:46 PM 
continuous at 0  summerset353  Real Analysis  2  February 22nd, 2010 03:37 PM 
continuous at any point iff continuous at origin  babyRudin  Real Analysis  6  October 24th, 2008 12:58 AM 