# Limit proof, absolute value problem.

• June 2nd 2009, 09:29 PM
craigmain
Limit proof, absolute value problem.
Hi,

I am reading a proof (in Spivak) that asserts that.

Given $0 < |x - a| < \delta_1$ and $|f(x)-l|< \epsilon$, and given $0 < |x - a| < \delta_2$ then $|f(x)-m|< \epsilon$ that $l=m$.

We choose $\delta = min(\delta_1, \delta_2)$ and the mid point between the two limits for epsilon. $\epsilon = \frac{|l-m|}{2}$.

The proof starts thus:

$|l-m| = |l - f(x) + f(x) - m|$

$\le |l - f(x)| + |f(x) - m|$
$< \frac{|l-m|}{2} + \frac{|l-m|}{2}$

He continues to prove a contradiction...

I have a problem at this point. We are substituting epsilon back in which seems simple. but does $|l - f(x)| = | f(x) - l|$ for all possible values of any function. Is the sign reversible even when it's a function. This seems to be required to get to the last step and I am unconvinced that it always holds. Am I missing something else?

Thanks
Regards
Craig.
• June 2nd 2009, 10:41 PM
CaptainBlack
Quote:

Originally Posted by craigmain
I have a problem at this point. We are substituting epsilon back in which seems simple. but does $|l - f(x)| = | f(x) - l|$ for all possible values of any function. Is the sign reversible even when it's a function. This seems to be required to get to the last step and I am unconvinced that it always holds. Am I missing something else?

For any particular $x$ in the domain of $f$ $f(x)$ is a number and so $|l-f(x)|=|f(x)-l|$.

But true for any particular $x$ in the domain of $f$ means the same as for all $x$ in the domain of $f$.

CB