Proving the Non-Continuity of f(x) at x=0

[lep11]

Homework Statement

Can f be defined to be continuous at x=0?

f(x)=x/(abs(x-1)-abs(x+1))

Prove why/why not.

Homework Equations

The Attempt at a Solution

The function is not defined at x=0 because f(0)=[0/0]. Therefore it cannot be continuous at x=0 but it is indeed continuous in its domain. How do I prove it cannot be continuous at x=0?

 

[pasmith]

Can you see how to choose c such that <br /> f(x) = \begin{cases} <br /> c &amp; x = 0, \\<br /> \frac{x}{|1 - x| - |x + 1|} &amp; x \neq 0<br /> \end{cases} is continuous at zero? (Hint: look at 0 &lt; |x| &lt; 1.)

 

[HallsofIvy]

Do you know the definition of continuous? If you do this is easy.

A function, f, is said to be "continuous at x= a" if and only if
1) f(a) exists
2) \lim_{x\to a} f(x)
3) \lim_{x\to a} f(x)= f(a)

(Since (3) obviously requires (1) and (2), often only three is stated.)

Now the problem does NOT ask whether or not f is continuous at x= 0 (It is obvious that it is not since f(0)- as you imply 0 is not even in the domain. The problem asks "Can f be defined to be continuous at x=0?" That is, can you assign a value to f(0) so that it is continuous?

That only affects (1) so you need to think about (2). If the limit, \lim_{x\to 0} f(x), does NOT exist then no value of f at 0 will make it continuous. If it does then it is clear what f(0) must be.

 

[lep11]

c= -½ But it's not the same function anymore?

 

[lep11]

I think I need to prove right-handed and left-handed limits are the same so limit at point x=0 exists. It's going to be a nightmare.

 

[Dick]

I think I need to prove right-handed and left-handed limits are the same so limit at point x=0 exists. It's going to be a nightmare.

It shouldn't be. Take x in the range (-1/2,0) or (0,1/2), then what sign are x-1 and x+1?

 

[lep11]

It shouldn't be. Take x in the range (-1/2,0) or (0,1/2), then what sign are x-1 and x+1?

Well, I have to prove using the definition, not "calculating" the limits.

 

[Dick]

Well, I have to prove using the definition, not "calculating" the limits.

You'll find that very easy once you actually try it.

 

[lep11]

Let 0< x <½ . Then | x/(|x-1|-|x+1|) -(-½)| = ½ | (2x+(|x-1|-|x+1|))/(|x-1|-|x+1|) | = ½ | (2x /(|x-1|-|x+1|) + 1 | = ½ | 2x /(|x-1|-x-1) +1| < | 2x /(|x-1|) +1|<...<...<epsilon whenever 0<x-0<½ ??

 

[Dick]

Let 0< x <½ . Then | x/(|x-1|-|x+1|) -(-½)| = ½ | (2x+(|x-1|-|x+1|))/(|x-1|-|x+1|) | = ½ | (2x /(|x-1|-|x+1|) + 1 | = ½ | 2x /(|x-1|-x-1) +1| < | 2x /(|x-1|) +1|<...<...<epsilon whenever 0<x-0<½ ??

If x is in (0,1/2) then x+1 is positive, so |x+1|=x+1. x-1 is negative, so |x-1|=(-(x-1))=1-x. Try writing your expression without the absolute values on each interval.

 

[lep11]

Let 0< x <½ .Then x/(|x-1|-|x+1|) = x/(-x+1-x-1)=-½ ?
Let -½<x<0 Then x/(|x-1|-|x+1|) = x/(-x+1-x-1)=-½ ?
However, that is not the epsilon, delta -definition.

 

[Dick]

Let 0< x <½ .Then x/(|x-1|-|x+1|) = x/(-x+1-x-1)=-½ ?
Let -½<x<0 Then x/(|x-1|-|x+1|) = x/(-x+1-x-1)=-½ ?
However, that is not the epsilon, delta -definition.

So once you define f(0)=(-1/2) then your function is the constant function f(x)=(-1/2) on that interval. Apply the epsilon delta definition to that. It's pretty easy.