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Homework Help: I've gone full retard stupid continuity question

  1. Sep 23, 2010 #1
    is the equation f(x)=(x^2-1)/(x+1) continuous?

    i know it can be reduced to f(x)=(x-1) but i remember that in doing so you divide by zero for x=-1 and thus it will be discontinuous at that point...


    i dunno i'm really tired tonight
     
  2. jcsd
  3. Sep 23, 2010 #2
    Never go full retard...

    The equation is discontinuous when the denominator is zero.
     
  4. Sep 23, 2010 #3
    thanks man. been a while since i had calc 1 i don't remember the exact rule of this situation. doesn't help that my high school calc teacher taught me a complete 180 from what my university professor did...
     
  5. Sep 24, 2010 #4
    Definition of continuity requires a function to be defined in point in which it is continuous.
     
    Last edited: Sep 24, 2010
  6. Sep 24, 2010 #5

    HallsofIvy

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    Science Advisor

    by the way, the problem is not to determine if the "equation" is continuous- it is to determine if the function defined by that equation is continuous. "continuity" is defined for functions, not equations.

    The definition of "f(x) is continuous at x= a" has three parts:
    1) That f(a) exist.
    2) That [itex]\displaytype \lim_{x\to a} f(x)[/itex] exist.
    3) That [itex]\displaytype \lim_{x\to a} f(x)= f(a)[/itex].

    As losiu99 says, [itex](x^2- 1)/(x- 1)[/itex] is not defined at x= 1 and so is not continuous there. [itex](x^2- 1)/(x-1)= x+ 1[/itex] for x not equal to 1 and is not defined at x= 1. Its graph is NOT the straight line y= x+ 1, it is the straight line y= x+ 1 with a hole at (1, 2).-
     
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