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Newbie

3 REPLIES 3

## 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. x2-3 becomes d2+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 d2+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 (x2-3)-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

## Finding delta

Here is another way.
Note that ½x2–3–1½=½x2–4½=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
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