# Epsilon Delta 2 - (1/x)

• Sep 13th 2013, 12:07 AM
mattjp213
Epsilon Delta 2 - (1/x)
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.
• Sep 13th 2013, 05:40 AM
HallsofIvy
Re: Epsilon Delta 2 - (1/x)
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 13th 2013, 12:51 PM
mattjp213
Re: Epsilon Delta 2 - (1/x)
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.