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Prove that, if f(x) is differentiable at x =c , then f(x) is continous at x=c

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    From the definition of the derivative, prove that, if f(x) is differentiable at x=c, then f(x) is continuous at x=c.

    2. Relevant equations

    f'(c) = lim [f(x)-f(c)]/(x-c) This is the definition for a function to be differentiable at
    x->c x=c.

    3. The attempt at a solution

    we are required to prove that
    lim f(x) = f(c) (this is what it means for the function to be continuous
    x->c

    lim f(x) - f(c) = 0
    x->c

    This looks alot like the numerator for the definition of differentiable at x=c.

    From here, i'm lost.
     
  2. jcsd
  3. Dec 6, 2011 #2
    Use the definition of limit to make an inequality (< ε for some ε > 0) to get yourself started and simplify. From there it is useful to change your x's into x[itex]_{n}[/itex]'s
     
  4. Dec 6, 2011 #3
    so right now i`m at the following:

    0 < |x-c| < δ --> |f(x)-f(c)| <ε

    But this thing has no numbers. I'm not used to it. I can perform the "reverse" triangle inequality on |x-c| but i'm still stuck. Could you give me 1 or 2 more step hint?
     
  5. Dec 6, 2011 #4

    vela

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    If you could show this was true, you'd be done. This statement says the limit of f(x) as x approaches c is f(c).
    You know the function is differentiable at x=c. That means, as you said earlier, that the limit
    [tex]\lim_{x \to c} \frac{f(x)-f(c)}{x-c}[/tex]exists. In terms of δ and ε, that means: given ε>0, there exists δ>0 such that |x-c| < δ implies
    [tex]\left|\frac{f(x)-f(c)}{x-c}-f'(c)\right| < \varepsilon[/tex]where f'(c) denotes the value of the limit. You need to somehow go from this statement to the statement: given ε'>0, there exists δ'>0 such that |x-c|<δ' implies |f(x)-f(c)|<ε'.
     
  6. Dec 6, 2011 #5

    vela

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    I wanted to add, it's probably not necessary to use the δ-ε definition of the limit for the proof. Just use the various properties of limits that you know.
     
  7. Dec 6, 2011 #6
    Thank you. I'll try it now.
     
  8. Dec 6, 2011 #7
    That'd probably be more likely on the exam. I don't think i'm expected to prove this with epsilon deltas anyway but even then, i'm sorta lost. I was hoping i didn't have to prove this with epsilon deltas as well. If we go back to where i'm at with the original post, what would be the next step in proving this with the limit laws?
     
  9. Dec 6, 2011 #8

    vela

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    In your first post, you noted that you want to show that
    [tex]\lim_{x \to c} f(x)-f(c) = 0.[/tex]You also noted it looked like the numerator in the definition of the derivative. So what if you looked at something like
    [tex]\lim_{x \to c} \left[\frac{f(x)-f(c)}{x-c}\cdot(x-c)\right]?[/tex]
     
  10. Dec 6, 2011 #9
    but here's where i got confused.

    if you did that, it's f'(c) = lim [f(x)-f(c)]
    x->c

    It still doesn't equal zero though.
     
  11. Dec 6, 2011 #10

    vela

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    How did you get f'(c) = lim [f(x)-f(c)]?
     
  12. Dec 6, 2011 #11
    oh wait nvm. Sorry but i meamnt to say that.

    lim f(x)-f(c)=0. We need that but do we know that f'(c) = 0?
    x->c
     
  13. Dec 6, 2011 #12
    You're trying to prove that [itex]\lim_{x \rightarrow c} f(x)-f(c) = 0[/itex], not that [itex]\lim_{x \rightarrow c} \frac{f(x)-f(c)}{x-c} = 0[/itex]. So you don't need to know that f'(c) is 0, you only have to know that it exists
     
  14. Dec 6, 2011 #13
    oh wait nvm. Sorry but i meamnt to say that

    We need that:

    lim f(x)-f(c)=0.
    x->c
    However we have:

    lim f(x)-f(c). It doesn't equal zero.
    x->c

    I'm just confused.
     
  15. Dec 6, 2011 #14
    Oh ok. So then the proof is basically done then right?
     
  16. Dec 6, 2011 #15

    vela

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    Depends. What did you say?
     
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