Epsilon Delta Proof

[MTL]

I worked and epsilon delta proof and got to a point where I knew it wasn't correct so then I just used trial and error to correctly pick the right delta, I will post my work, but I am really interested in the correct way to do problems similar to this.
Given: The limit of (2-(1/x)) as x approaches 1 is equal to 1. Prove that a delta exists for Epsilon equal to 0.1.
My Work:
By definition of delta: |x-1| < d
Property of equality: (1/x)*|x-1| < (1/x)*d
Distribute: |1-(1/x)| < d(1/x)
Substitution and commutative properties: |2-(1/x)-1| < d(1/x)
Here I realize that I have broken some rules of algebra so I just use trial and error to decide that the best value for x is (10/11).
Now, d(1/x) = d(11/10)
Substitution: d(11/10) = E
Prop. of Equality: d = E(10/11)
Sub. given info.: d = (1/10)*(10/11)
Now, d = (1/11) about 0.091 which is less than E = 0.1 and checks with the book.
How is this done the correct way?

 

[pka]

I worked and epsilon delta proof and got to a point where I knew it wasn't correct so then I just used trial and error to correctly pick the right delta, I will post my work, but I am really interested in the correct way to do problems similar to this.
Given: The limit of (2-(1/x)) as x approaches 1 is equal to 1. Prove that a delta exists for Epsilon equal to 0.1.

Start with \(\displaystyle |x-1|<0.5\) from which it follows that \(\displaystyle 0.5<x<1.5\) or \(\displaystyle \frac{1}{|x|}<2\)

Note that \(\displaystyle \left| {\left( {2 - \frac{1}{x}} \right) - 1} \right| = \frac{{\left| {x - 1} \right|}}{{\left| x \right|}} < 2\delta \) So pick \(\displaystyle \delta = \min \left\{ {\frac{\varepsilon}{2} ,0.5} \right\}\).

Now you can complete the proof.

 

[MTL]

Good Grief

Start with \(\displaystyle |x-1|<0.5\) from which it follows that \(\displaystyle 0.5<x<1.5\) or \(\displaystyle \frac{1}{|x|}<2\)

Note that \(\displaystyle \left| {\left( {2 - \frac{1}{x}} \right) - 1} \right| = \frac{{\left| {x - 1} \right|}}{{\left| x \right|}} < 2\delta \) So pick \(\displaystyle \delta = \min \left\{ {\frac{\varepsilon}{2} ,0.5} \right\}\).

Now you can complete the proof.

So after your hint and an hour and a half I figured it out, and I am replying before I look at the next question. The whole chapter they use polynomials as examples then the first E-d proof they lay out is that one!!!!!!! Good news is I shouldn't forget that anytime soon.

Given f(x) = 2-(1/x). Find d such that if 0 < |x-1| < d then |f(x)-1| < 0.1.

By def. of a limit and given info.: |(2-(1/x)) -1| < 0.1
Combining like terms and rules for adding fractions: |(x-1)/x| < 0.1
Properties of Abs. Value: |x-1|/|x| < 0.1
Def. of a limit and given: |x-1| < d
Properties of Abs. V.: 1-d < x < d+1
Reciprocal Prop.: 1/|x| < 1/(1-d)
Proven previously: |x-1|*(1/|x|) < 0.1
Substitute: d*(1/(1-d)) = 0.1
Prop. of Algebra: d = 1/11
I then ran it from the delta definition back to the epsilon using 1/11 for delta and it worked beautifully. Thank you so much for your help. I need to strengthen my knowledge of properties of abs. value and use this tool of comparing inequality to inequality.