Work backwards.
So we want the maximum such that . This is clearly .
When proving that limit (x->5) 1/x = 0.2, find the largestδ > 0 that works for ε = 0.0025 = 1/400
My attempt: | 1/x - 1/5 | = | (5-x)/5x | = | (x-5) /5x | < ε. | (x-5) /5x | < 1/400 whenever 0 < | x-5| < δ. So for some real number M, if 1/|5x| < M, then | (x-5) /5x | < M |x-5|. Hence | x-5| < ε/M = δ.
Now |x-5| < 1 implies 4<x<6 which further implies 1/30 < 1/|5x| < 1/20, thus M= 1/20 is suitable. So we can take δ = min{1,20*ε}. That is the largest δ is 20/400 = 0.05.
Is this argument correct?
Hello, pessimist92. Your procedure is correct if you want to construct an epsilon-delta proof. It gives you a formula so that for any given epsilon, you can construct a delta that works. But the question asked for the *largest* delta that works for this particular epsilon, so you need to do what Nehushtan says. Your formula gives , which is less than .
- Hollywood