Epsilon Delta 2 - (1/x)

Sep 2013
10
0
Michigan
I have been working at epsilon delta problems for hours each day and I can do many of them, but this one is destroying me and I'm tapping out for help now.

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.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Do you understand that there are an infinite number of "correct" answers? If some "d" works then so does any smaller number.

You are correct that we are led to |x- 1|< |x|e. Now, since we are looking for "x close to 1" we might require that x> 1/2. In that case |x|e> (1/2)e so that if |x- 1|< (1/2)e it is certainly less than |x|e. And taking e= 0.1, we have |x- 1|< d= (1/2)(0.1)= 0.05.

Or we might require that x> 2/3. In that case |x|e> (2/3)e so that if |x- 1|< (2/3)e it is certainly less than |x|e. And taking e= 0.1, we have |x- 1|< d= (2/3)(0.1)= 0.066666....
 
Sep 2013
10
0
Michigan
Thanks for letting me know that my work wasn't incorrect. I realize that there are many possible d values, but didn't understand why the book had a different answer. I finally discovered that they were using f(x) + or - e to solve for x and plugging each possibility into the |x-1| < d inequality. When f(L-e) = 2-(1/x) then x = 10/11. |(10/11)-1| = 1/11 which is the books answer.

I appreciate the verification of my answer. Now I know I am still on the right track.