Question

Epsilon-Delta Limits, anyone? The concept goes over my head.

Answer 1

Do you have a certain problem ?

Answer 2

Yes! Lemme type it out for you.

Answer 3

Find δ such that 0 < |x - 1 | <δ then | f(x) - 1 | < 0.1 for the function f(x) = 2 -1/x

Answer 4

Well,first of all you can watch this video to get an intution on epilson delta limits

- https://www.khanacademy.org/math/calculus/limits_topic/epsilon_delta/v/epsilon-delta-limit-definition-1#!

Answer 5

Took a screen capture of the problem / graph; maybe it will help.

I have seen that video, but I'll give it another watch.

Answer 6

What we want is for \[ | f(x)-1| = \left|\left(2-\frac{1}{x}\right)-1\right| = \left|1-\frac{1}{x}\right|<0.1\] by an appropriate bound on x, or really, x-1. That means \[ -0.1<1-\frac{1}{x}<0.1\] and so \[-1.1<-\frac{1}{x}<-0.9\] and so \[0.9<\frac{1}{x}<1.1\] Now we may want to worry about x being 0, so if we ensure that delta is less than 1, then it will make sure that x is not 0 on that interval.

Now by a little piece of Algebraic trickery, this last inequation implies \[\frac{1}{1.1}<x<\frac{1}{0.9}\] and therefore \[\frac{1}{1.1}-1<x-1<\frac{1}{0.9}-1\]

With this you should be able to find bounds on \[|x-1|\] which ensures that this is true.