- #1
ltjrpliskin
- 13
- 0
f(x) = x^2 and we want to show that as x -> a, f(x) -> a^2.
So, we must ensure that |x^2 - a^2| < eps.
The book (Calculus by Spivak) starts of by bounding |x - a| < 1 which leads us to
|x + a| < 2|a| + 1.
This is the part where I get lost:
I can understand that we require |x - a| < eps/ (2|a| + 1),
but
why do we choose |x - a| < min(1, eps/ (2|a| + 1)) ?
And what happens if min(1, eps/ (2|a| + 1)) = 1 ?
Thank you very much for your help in advance! :)
So, we must ensure that |x^2 - a^2| < eps.
The book (Calculus by Spivak) starts of by bounding |x - a| < 1 which leads us to
|x + a| < 2|a| + 1.
This is the part where I get lost:
I can understand that we require |x - a| < eps/ (2|a| + 1),
but
why do we choose |x - a| < min(1, eps/ (2|a| + 1)) ?
And what happens if min(1, eps/ (2|a| + 1)) = 1 ?
Thank you very much for your help in advance! :)