What is the reasoning behind bounding |2/3x| instead of |3x/2| in this limit?

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SUMMARY

This discussion focuses on the reasoning behind bounding |2/3x| instead of |3x/2| when evaluating limits in calculus. The key point is that bounding |x+3| is necessary when proving limits at x=3, as it ensures |x-3| remains less than a constant. The equivalence of |2/(3x)||x-1/2|< A to |x-1/2|< A/|2/(3x)| is established, clarifying the relationship between the bounds. The participants emphasize the importance of understanding the context of the limit being evaluated.

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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| ?
 
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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|?
 
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 there's no chance of getting a zero on the bottom line of the right hand side
 

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