# Thread: why they chose maximum and not minimum functions..

1. ## why they chose maximum and not minimum functions..

it uses minimum function because we need to squeeze the function
as close as possible to the limit(example 4).
http://www.mathhelpforum.com/math-he...ta-proofs.html

but i got this proove that use maximum.
http://img440.imageshack.us/img440/9160/25023372lz4.gif

why?

2. Originally Posted by transgalactic
it uses minimum function because we need to squeeze the function
as close as possible to the limit(example 4).
http://www.mathhelpforum.com/math-he...ta-proofs.html

but i got this proove that use maximum.
http://img440.imageshack.us/img440/9160/25023372lz4.gif

why?
Im sorry, maybe it is just my computer but I cannot see the majority of the paper. It might serve you better to start posting in LaTeX. From what I can see seems as though you have a case where you have two inequalities and you want them both to be true, so if $x and $y then if I have a function that is sometimes $x$ and sometimes $y$ then the only thing we can say the function is always less than is $\max\left\{x,y\right\}$.

For example, suppose that $g(x) and $h(x) then the function $f:x\longmapsto\left\{\begin{array}{rcl} h(x) & \mbox{if} & x\in\mathbb{Q}\\ g(x) & \mbox{if} & x\not\in\mathbb{Q}\end{array}\right.$ then on any interval we can really only say that $f<\max\left\{M,N\right\}$.

3. then why in the article they use minimum function?

4. Originally Posted by transgalactic
then why in the article they use minimum function?
In one they are working with |x-a| in the other they are looking at $\frac{1}{|x-a|}$. The reciprocal swaps "max" and "min".

5. i understand the because its 1/(expression) then it flips

why in the first we choose min(the contained bound,epsilon/contained bound)

??

6. in the third example of the article we do not use 1/|x-a| expression

so why they use min and not max ??