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....

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.