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October 12th, 2009, 08:14 PM  #1 
Senior Member Joined: Apr 2009 Posts: 201 Thanks: 0  counter example to a skewed deltaepsilon defns?
can you check this please? disprove: for any delta, there exists an epsilon such that 0 <  xa < delta > f(x)  L  < epsilon this is just the negative of the usual deltaepsilon definition, so I can show how there's at least one case where there isn't an epsilon for ANY delta... for any epsilon, there exists a delta such that f(x)  L  < epsilon > 0 <  xa < delta I picked a periodic function like sin(x) or cos(x) where you can find f(x) = f(x1) = f(x2) = f(x3).. et c. so, f of any of these x's  L should be less than epsilon, but if you then take one of those x's x some fixed a is not within delta  since you can pick an a from a different period such that f(x)  L < epsilon but xa is not less than delta \ edit: nevermind, you could just pick delta to be as big as you want, so this wouldn't work thanks 
October 13th, 2009, 05:53 AM  #2  
Senior Member Joined: Dec 2008 Posts: 306 Thanks: 0  Re: counter example to a skewed deltaepsilon defns?
Are you trying to show that if this Quote:
 
October 13th, 2009, 07:59 AM  #3  
Senior Member Joined: Apr 2009 Posts: 201 Thanks: 0  Re: counter example to a skewed deltaepsilon defns? Quote:
I need to find an example where a limit would exist but this definition wouldn't be a proper definition thanks  
October 13th, 2009, 09:10 AM  #4  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: counter example to a skewed deltaepsilon defns? Quote:
Try to find a discontinuous function which satisfies this property. Quote:
This doesn't work for the first example, because you can choose any epsilon > 2, and it works (all it says is there exists an epsilon it says nothing about whether this epsilon is large or small)  
October 13th, 2009, 11:50 AM  #5  
Senior Member Joined: Apr 2009 Posts: 201 Thanks: 0  Re: counter example to a skewed deltaepsilon defns? Quote:
I will think more about the first case thanks for the help! you're always helping me, I appreciate it  
October 13th, 2009, 12:03 PM  #6  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: counter example to a skewed deltaepsilon defns? Quote:
 
October 14th, 2009, 05:27 PM  #7 
Senior Member Joined: Apr 2009 Posts: 201 Thanks: 0  Re: counter example to a skewed deltaepsilon defns?
Hi, I think for the first part, if you have a discontinuous function that has very large intervals between each point or segment, you wouldn't be able to find an epsilon for a certain delta because the epsilon would have to span infinitely

October 15th, 2009, 03:36 AM  #8  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: counter example to a skewed deltaepsilon defns? Quote:
That doesn't matter though: My suggestion was to look at a discontinuous function which satisfies the statement you gave. If the function is discontinuous, and doesn't satisfy the sentence, you get nothing, because you are trying to show that this definition doesn't line up with continuity. In other words, call Quote:
But... Your suggestion helped clarify the solution in my head Look at the floor function. It is not continuous at any integer (convince yourself). So let a be an integer and show that f is P at a. Then you have a point for which floor(a) satisfies P at a, but floor(a) is not continuous at a; in other words, the definitions are not the same.  
October 15th, 2009, 02:20 PM  #9  
Senior Member Joined: Apr 2009 Posts: 201 Thanks: 0  Re: counter example to a skewed deltaepsilon defns? Quote:
but if you have discontinuity, the limit doesn't exist anyway, you would have to define single sided limits, wouldn't you? I'm thinking that the definition would be satisfied for the floor function, but since x>a takes on different limits, it can't be true (even though it is satisfied..) thanks  
October 15th, 2009, 10:00 PM  #10 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: counter example to a skewed deltaepsilon defns?
I thought the problem was "Show that the statement "for any delta, there exists an epsilon such that 0 <  xa < delta > f(x)  L  < epsilon" is not equivalent to the definition of continuity." You have two options: 1) Show a function which is continuous, but does not satisfy the property above OR 2) Show a function which satisfies the property above, but is not continuous. I *think* functions that fit option 1 ore more degenerate than ones that fit option 2... I could be wrong though, and there might be a very nice function which is continuous but does not satisfy the statement. The floor function is onesided continuous, but onesided continuity is not continuity: it's onesided continuity. So, the floor function is not continuous at any integer. If you can show that the floor function satisfies the above sentence, you've done number 2. 

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