1. Feb 7, 2008 #1

   coverband

   

   If |x-3| < A/|x+3| we need to bound |x+3| right?
   
   Now if you take |2/3x||x-1/2| < A why do we bound |2/3x| and not |3x/2| ?

    

   coverband, Feb 7, 2008
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3. Feb 7, 2008 #2

   HallsofIvy

   

   Staff Emeritus

   Science Advisor

   Whether you need to "bound |x+3|" depends on what you are doing! If you are trying to prove that a certain limit is true at x= 3, then, yes, since you want |x-3|< constant, you are going to need a bound on |x+3|.
   
   As for |2/(3x)||x- 1/2|< A, that is the same as |x- 1/2|< A/|2/(3x)|. Comparing with you inequality above, yes, the thing corresponding to |x+3| is |2/(3x)|. Why would you think it would be |3x/2|?

    

   HallsofIvy, Feb 7, 2008
4. Feb 7, 2008 #3

   coverband

   

   Limits:
   
   0<|x-(1/2)|<B implies |f(x)-L|<A
   
   If we arrive at
   
   |2/3x||x-(1/2)| < A, I thought we could just write this as |x-(1/2)| < A |3x/2|
   and not worry about bounding anything coz theres no chance of getting a zero on the bottom line of the right hand side

    

   coverband, Feb 7, 2008

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