
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  =  (5x)/5x  =  (x5) /5x  < ε.  (x5) /5x  < 1/400 whenever 0 <  x5 < δ. So for some real number M, if 1/5x < M, then  (x5) /5x  < M x5. Hence  x5 < ε/M = δ.
Now x5 < 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?

Re: Epsilondelta proof
Work backwards.
$\displaystyle \frac1{400}<\frac1x\frac15<\frac1{400}$ $\displaystyle \iff$ $\displaystyle \frac{79}{400}<\frac1x<\frac{81}{400}$ $\displaystyle \iff$ $\displaystyle \frac{400}{81}<x<\frac{400}{79}$ $\displaystyle \iff$ $\displaystyle \frac5{81}<x5<\frac5{79}$
So we want the maximum $\displaystyle \delta$ such that $\displaystyle \delta<x5<\delta\ \Rightarrow\ \frac5{81}<x5<\frac5{79}$. This is clearly $\displaystyle \delta=\frac5{81}$.

Re: Epsilon Delta Proof
Hello, pessimist92. Your procedure is correct if you want to construct an epsilondelta 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

Re: Epsilon Delta Proof
Thanks to both of you. Now I understand the logic of these type of Epsilondelta questions.