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09-26-2003
03:00 AM

09-26-2003
03:00 AM

Finding delta

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09-26-2003
03:00 AM

09-26-2003
03:00 AM

Finding delta

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09-26-2003
03:00 AM

09-26-2003
03:00 AM

Finding delta

You want the limit as x->2, so you are concerned with x of the form 2+d for small values of d. The question is how small.

Start by substituting for x. x^{2}-3 becomes d^{2}+4d+1. Since we are interested in the difference of this from one, we can subtract one and see how small d must be to make what's left small enough.

Subtracting one leaves us with d^{2}+4d. And that must be made smaller than e in magnitude. We can factor this as d(d+4). If d is greater than zero, then clearly this difference is also greater than zero and we need an upper limit for it. Further if d<1 then d(d+4)<5d. And if d<e/5 then 5d<e

So if d is less than both one and e/5 we will have (x^{2}-3)-1 less than e. You can work out a similar set of inequalities for d (and the difference) less than zero, and combine them to get inequalities in the absolute values.

Tom Gutman

Start by substituting for x. x

Subtracting one leaves us with d

So if d is less than both one and e/5 we will have (x

Tom Gutman

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09-26-2003
03:00 AM

09-26-2003
03:00 AM

Finding delta

Here is another way.

Note that ½x^{2}–3–1½=½x^{2}–4½=½x–2½½x+2½. If e >0 then take d =min{1,e/5}.

Now, if ½x–2½<d then ½x+2½£½x–2½+½4½£5. Now you can put this togther.

PKA

Note that ½x

Now, if ½x–2½<d then ½x+2½£½x–2½+½4½£5. Now you can put this togther.

PKA