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One sided limits

  1. Nov 9, 2005 #1
    Hi people,
    could anyone tell me how to prove that the limit as f(x) approaches a from above equals the limit as f(x) approaches a from below? I can't see how to approach this proof, thx

    Jack
     
  2. jcsd
  3. Nov 9, 2005 #2

    mathman

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    The question as you stated it is too vague. In any case if f(x) is discontinuous, it just won't be true.
     
  4. Nov 9, 2005 #3
    Unless you have some sort of piecewise function, I see this as fairly straightfoward. If

    (1) [tex]\lim_{x\rightarrow{a+}}f(x)=f(a)=\lim_{x\rightarrow{a-}}f(x)[/tex],

    then f(x) is continuous at x=a. What is your proof concerning? Continuity, delta-epsilon proofs?
     
  5. Nov 9, 2005 #4
    i'm guessing it's the epsilon-delta stuff.
     
  6. Nov 10, 2005 #5
    That condition is not accurate, consider this function:

    [tex]f(x) = \frac{x^2-1}{x-1}[/tex]

    The limit above and below f(1) is equal to 2 though it is undefined at that point.
     
  7. Nov 10, 2005 #6
    It's obvious that Jameson meant for f(a) (a = 1 in this case) to exist, seeing as he mentioned that something should be equal to it, and in that case, the condition is accurate.
     
  8. Nov 13, 2005 #7
    I know what he means. He is talking about a function between points (a,f(a)) and (b,f(b)) and he wants to know how to prove the limit at x=a or x=b

    From what I remember in Chapter 2, Calculus AB all you need to do is see limit as x->a or b->b from the existent side, and then plug in the value into the function. If it's the same, it's continuous. If it's not, no continuity.

    It's like the following:

    limit as x->a of f(x)=b
    and f(a)=b

    makes a function continuous at point (a,f(a))

    right? It's been a good few months.
     
  9. Nov 13, 2005 #8
    Yes, I was stating the conditions for a function being continuous at the point a. However, I was not saying continuity is necessary for a limit to exist. Sorry if I was unclear.
     
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