Expert: Ahmed Salami - 11/10/2008

Question
I was asked to find a delta such that:
  abs( 1/(x-1) -1 ) < .01  when
             0<abs(x - 2) < delta
the epsilon is .01 and I have to find delta.

 I get as far as      -.01 <  (2-x)/(x-1) < .01

 but the fraction has me puzzled because I don't know if
 (x-1) or (2-x) are positive or negative or what?  

Answer
Hi Bernadette,
|1/(x-1) - 1|
= |(2-x)/(x-1)|
= |(x-2)/(1-x)| = |(x-2)/(x-1)|   since |A| = |-A|
= |x-2|/|x-1|
So we have
|x-2|/|x-1| < 0.01
If we take d < 1 and require that |x-2| < d, then we have
|x-2| < 1
i.e -1 < x-2 < 1
adding 1 to all parts give us
0 < x-1 < 2
or |x-1| < 2
Therefore,
|x-2|/|x-1| > |x-2|/2
but |x-2|/|x-1| < 0.01, and so it follows that
|x-2|/2 < 0.01
|x-2| < 0.02
We can take our d to be 0.02

Regards.