Problem with procedure for solving an epsilon delta problem if $|x-2|<\delta$ then $|(1/x)-0.5|<0.2$
my procedure I’ve come up with by seeing it work 3 times on hw:
1. change the expression from looking like |…|<a to -a<…<a.
2. get x alone in the middle.
3. set left sides of both expressions equal to each other and same for right sides.
4. solve for delta.
5. take the smaller of the 2 delta values.

This doesn’t work for problem above I get -4/3 and -4/7 for delta values and I put -4/3 into answer and it came out as wrong. I need a modification for the above procedure.
Edit
How i got the values -4/3 and -4/7:


*

*$-\delta<x-2<\delta$ and $-1/5<(1/x)-(1/2)<1/5$

*$2-\delta<x<\delta+2$ and   $10/3<x<10/7$

*$2-\delta=10/3$ and $\delta+2=10/7$

*$\delta=-4/3$ and $\delta=-4/7$

*$-4/3<-4/7$ -> $-4/3$

 A: $\delta$ has to be positive!  You don't say how you got those negative values so it is impossible to say what you might have done wrong.
You want to find $\delta$ such that if $|x- 2|< \delta$ then $|(1/x)- 0.5|< 0.2$.  As your "procedure" says, write that second inequality as $-0.2< (1/x)- 0.5< .2$.  Now "solve for x".  Add 0.5 to both sides to get $0.3< 1/x< 0.7$.  Since $0.3< 1/x$, x is clearly positive so we can invert each part but also remember that this will reverse the inequality: $(1/0.7)< x< 1/0.3$.
(A little more rigorously: if $0.3< 1/x$ then $0.3x< 1$ so $x< (1/0.3)$.  If $1/x< 0.7$ and x is positive, then $1< 0.7x$ so $(1/0.7)< x$.)
For the other inequality, $|x- 2|< \delta$, we have $-\delta< x- 2< \delta$ so $2-\delta< x< 2+ \delta$.  
Now, setting "left sides of both expressions equal to each other and same for right side" we have $1/0.7= 1.428= 2-\delta$ so $\delta= 2+ 1.428= 3.428$ and $1/0.3= 3.333= 2+ \delta$ so $\delta= 3.333- 2= 1.333$.  The smaller of those is 1.333 or 4/3.
