1. ## 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?

2. ## Re: Epsilon-delta proof

Work backwards.

$\displaystyle -\frac1{400}<\frac1x-\frac15<\frac1{400}$ $\displaystyle \iff$ $\displaystyle \frac{79}{400}<\frac1x<\frac{81}{400}$ $\displaystyle \iff$ $\displaystyle \frac{400}{81}<x<\frac{400}{79}$ $\displaystyle \iff$ $\displaystyle -\frac5{81}<x-5<\frac5{79}$

So we want the maximum $\displaystyle \delta$ such that $\displaystyle -\delta<x-5<\delta\ \Rightarrow\ -\frac5{81}<x-5<\frac5{79}$. This is clearly $\displaystyle \delta=\frac5{81}$.

3. ## 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 $\displaystyle \delta=\frac{1}{20}$, which is less than $\displaystyle \frac{5}{81}$.

- Hollywood

4. ## Re: Epsilon Delta Proof

Thanks to both of you. Now I understand the logic of these type of Epsilon-delta questions.