# |2/3x||x-(1/2)| < A

1. Feb 7, 2008

### 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| ?

2. Feb 7, 2008

### HallsofIvy

Staff Emeritus
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|?

3. Feb 7, 2008

### 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

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