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Does this limit exist?

  1. Jun 25, 2013 #1
    This has confused me a bit
    I'm typing on my phone so I can't use math symbols sorry
    F(x) = 1 ; x=5
    0 otherwise (x=/5)
    Does lim X approaches 5 f(x) exists and = 1(f continuous at 5) ? Or does it not(because when we approach a were also in the x=/a space so there is also 1)
    can someone be really kind and Prove it using epsilon?
    I know you can just put alpha = any number you want since 0<epsilon always
    I think I'm confusing these functions with the ones like sinx..
     
    Last edited: Jun 25, 2013
  2. jcsd
  3. Jun 25, 2013 #2

    haruspex

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    Not at all sure what you mean by that last statement. When x ≠ a, F(x) = 0, yes? What is F(1/n), n = 1, 2, ...?
     
  4. Jun 25, 2013 #3
    I'm sorry forgot to set the integer fixed
     
  5. Jun 25, 2013 #4
    F(x) : R--) R
    0 x=5
    1 x=/5
    Is f continuous at 0?
     
  6. Jun 25, 2013 #5

    Ray Vickson

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    What is the *definition* for f(x) to be continuous at x = 0? Does your f satisfy that definition? Show your work!
     
  7. Jun 25, 2013 #6

    lurflurf

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    The point of this exercise is to make the point that the value of a function at the value of the limit does not matter.
    That is

    $$\lim _{x \rightarrow a} \mathop{f}(x) $$
    Does not depend on the value of f(a)

    Perhaps there is a function G that you know of such that
    G(x)=A if x=5
    G(x)=F(x) otherwise (x=/5)
    Choose G(5) whatever you like. So that you know the limit.
     
  8. Jun 26, 2013 #7
    I'm sorry but I had a huge confusion with this, I'm not sure if when it approaches 5 is it f(5) or not..
    I used epsilon delta but. I can't decide what f(x) should I be using
    lx-5l<alpha. lf(x)-f(5)l< epsilon..
    What f(x) should I use
    Just to mention this is not my first time with limits but out of nowhere I found myself unable to do this..
     
  9. Jun 26, 2013 #8
    What are you saying is that the limit of my function x approaches 5 is. 1 and not 0?
    We can choose g(5)=100 that doesn't mean limit x approaches 5 is 100?
     
  10. Jun 26, 2013 #9
    thanks everyone i've got it the limit = 1 => f is not continious at 5
    this whole problem was caused by an exercise @ spivak's calculus it was f is discon tinious at l but i thought it is continious .
     
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