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 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. |x-c|
February 7th, 2012, 06:00 AM   #2
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Re: continuouse for 1/x

Quote:
 Originally Posted by frankpupu holds for all points c and x in (0, infinity)
Should be (0, 1), no?

 February 9th, 2012, 05:43 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,390 Thanks: 2015 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.

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