# sups and lim

• Apr 11th 2013, 11:46 PM
orir
sups and lim
f is a continous function in (0,1) which maintains lim x-->0+ f(x)=-1, and lim x-->1- f(x)=1
let's define: A={x in (0,1)| f(x)=0}

s=supA. i need to prove that f(s)=0.

well, i know that in oter words i need to prove that s is part of the group A, and is its maximum. But i don;t know how to proceed from that, because
my proffesor isn't satisfied with verbal proof. he wants formal mathematical proof [i saw him proving existence of sup using the definition of the limit (epsilon-delta) and by that contradicted the existence of another upper-bound number which smaller than the one we needed to prove]
• Apr 12th 2013, 03:07 AM
Plato
Re: sups and lim
Quote:

Originally Posted by orir
f is a continous function in (0,1) which maintains lim x-->0+ f(x)=-1, and lim x-->1- f(x)=1
let's define: A={x in (0,1)| f(x)=0}, s=supA. i need to prove that f(s)=0.

There two cases: 1) $s\in A$ and 2) $s\notin A$.

Case 1) is trivial.

In case 2), there must be a sequence $(a_n)\subset A$ such that $(a_n)\to s$.
But $f$ is continuous, so what can we say about $f(a_n)\to~?$
• Apr 12th 2013, 03:25 AM
orir
Re: sups and lim
well, we didn't learn sequences in calculus yet...
is there any other way to solve this without using sequences??
• Apr 12th 2013, 03:51 AM
Plato
Re: sups and lim
Quote:

Originally Posted by orir
well, we didn't learn sequences in calculus yet...
is there any other way to solve this without using sequences??

Yes there are other ways. But they are tedious. I will not do it.

You can. If $f(s)\ne 0$ then $(\exists \epsilon>0)(\forall x\in(s-\epsilon,s+\epsilon)[f(x)\ne 0]$.