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Is function continuous@x=0?

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Can f be defined to be continuous at x=0?

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

    Prove why/why not.

    2. Relevant equations


    3. 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?
     
    Last edited: Feb 22, 2015
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  3. Feb 22, 2015 #2

    pasmith

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    Can you see how to choose [itex]c[/itex] such that [tex]
    f(x) = \begin{cases}
    c & x = 0, \\
    \frac{x}{|1 - x| - |x + 1|} & x \neq 0
    \end{cases}[/tex] is continuous at zero? (Hint: look at [itex]0 < |x| < 1[/itex].)
     
  4. Feb 22, 2015 #3

    HallsofIvy

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    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) [itex]\lim_{x\to a} f(x)[/itex]
    3) [itex]\lim_{x\to a} f(x)= f(a)[/itex]

    (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, [tex]\lim_{x\to 0} f(x)[/tex], 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.
     
    Last edited by a moderator: Feb 22, 2015
  5. Feb 22, 2015 #4
    c= -½ But it's not the same function anymore?
     
    Last edited: Feb 22, 2015
  6. Feb 22, 2015 #5
    I think I need to prove right-handed and left-handed limits are the same so limit at point x=0 exists. It's gonna be a nightmare.
     
  7. Feb 22, 2015 #6

    Dick

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    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?
     
  8. Feb 22, 2015 #7
    Well, I have to prove using the definition, not "calculating" the limits.
     
  9. Feb 22, 2015 #8

    Dick

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    You'll find that very easy once you actually try it.
     
  10. Feb 22, 2015 #9
    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 ??
     
  11. Feb 22, 2015 #10

    Dick

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    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.
     
  12. Feb 22, 2015 #11
    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.
     
    Last edited: Feb 22, 2015
  13. Feb 22, 2015 #12

    Dick

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    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.
     
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