1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper

    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 .
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted