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

My work is this:

|2-(1/x)-1| = |1-(1/x)| = |(x/x)-(1/x)| = |x-1|/|x| < e , so |x-1| < |x|e

I then bound d to find the minimum value of x.

d < 1 so, 0 < |x-1| < 1 = -1 < x-1 <1 = 0 < x < 2 , 0e = 0 and d > 0 so we must choose x's value to be 2.

If we allow d = min{1, 2e}, and if 0 < |x-1| < d, then |2-(1/x)-1| = |x-1|/|x| < e , 2e/2 = e

The question asks for d if e = 0.1. My d is 2e so d would be 2(0.1) = 0.2.

I thought I did this correctly, but the answer in the back of my book says d = 1/11 or ~0.091.

After retrying this problem many times over the past week, I still cannot figure this out. Please teach me how to solve this problem.