Epsilon - Delta ProofsDate: 1/23/96 at 13:47:58 From: Catherine Miller Subject: Def. of Limit? lim f(x) = L x-->c 0<|x-c|< d (delta), then |f(x)-L|< E (epsilon) Example: lim(2x - 5) = 1 find delta such that |(2x - 5) - 1| < 0.01 whenever 0 < |x - 3| < delta the f(x) / \the L ?/ \the c I'm confused about the f(x) in the epsilon function and the x in the delta function. Does the delta (x) become a value when solving these equations? Date: 6/20/96 at 9:56:40 From: Doctor Jerry Subject: Re: Def. of Limit? To show that the the limit of 2x-5 is 1 as x approaches 3, you must show that for each choice of E you can find a delta so that |(2x - 5) - 1| < E whenever 0 < |x - 3| < delta. With your choice of E=0.01, let's find a suitable delta by doing some scratch work first. We want |(2x - 5) - 1| < 0.01. Rewrite this as |2x - 6| < 0.01 and, again, as |x - 3| < 0.01/2. In this form, it appears that if we take delta =0.01/2, then |(2x - 5) - 1| < 0.01. In fact, if you replace 0.01 by E, you can in the same way, obtain a general procedure for choosing delta, once given an E. Namely, choose delta=E/2. I hope this answers your question. If not, write back to us. -Doctor Jerry, The Math Forum |
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