# Epsilon Delta Proof

• February 28th 2013, 12:18 AM
pessimist92
Epsilon Delta Proof
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?
• March 2nd 2013, 05:14 AM
Nehushtan
Re: Epsilon-delta proof
Work backwards.

$-\frac1{400}<\frac1x-\frac15<\frac1{400}$ $\iff$ $\frac{79}{400}<\frac1x<\frac{81}{400}$ $\iff$ $\frac{400}{81} $\iff$ $-\frac5{81}

So we want the maximum $\delta$ such that $-\delta. This is clearly $\delta=\frac5{81}$.
• March 2nd 2013, 10:00 AM
hollywood
Re: Epsilon Delta Proof
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 $\delta=\frac{1}{20}$, which is less than $\frac{5}{81}$.

- Hollywood
• March 3rd 2013, 11:24 PM
pessimist92
Re: Epsilon Delta Proof
Thanks to both of you. Now I understand the logic of these type of Epsilon-delta questions.