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Math Help - continuity question

  1. #1
    Junior Member
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    continuity question

    Regarding
    <br />
   f(x) = \left\{<br />
     \begin{array}{lr}<br />
         x^2 & : x < 0\\<br />
       0 & : 0 \leq x \leq 2<br />
     \end{array}<br />
   \right.<br />
    Is f continuous on [0,1] or [1,2]?
    and is f 'right continuous' on [0,1]?
    and why?

    [I dont really know what it means for a function to be 'right continuous'...
    any help on this?]
    Last edited by binkypoo; October 23rd 2009 at 11:34 AM.
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  2. #2
    Senior Member slevvio's Avatar
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    I think youve made a mistake in that, I think it should say  f(x) = x^2 for x < 0.

    If a function is right continuous then the limit of f(x) as you approach a from the right is f(a).

    i.e.

     \forall \epsilon > 0, \exists \delta > 0 such that  \forall x \in (a, a+\delta), |f(x) - f(a) | < \epsilon

    There's some pictures of these functions on wikipedia
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  3. #3
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    Note that for a function to be "continuous" at a point, it must be both "right continuous" and "left continuous".

    What was your answer to the question "Is f continuous on [0,1] or [1,2]?"
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  4. #4
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    I believe f is not continuous on [0,1] because the right side limit and left side limit at 0 are not equal.
    on [1,2] it seems to me that it is continuous but Im not sure how to prove either of these claims
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  5. #5
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    Quote Originally Posted by binkypoo View Post
    I believe f is not continuous on [0,1] because the right side limit and left side limit at 0 are not equal.
    The "right side limit" is \lim_{x\to 0^+} f(x)= \lim_{x\to 0} 0 and the "left side limit" is \lim_{x\to 0^-} f(x)= \lim_{x\to 0} x^2.

    What are those two limits?

    on [1,2] it seems to me that it is continuous but Im not sure how to prove either of these claims
    on [1, 2], f(x) is identically 0. Can't you prove that \lim_{x\to x_0} 0= 0?
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  6. #6
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    Sorry but I just realized I wrote the function wrong, here is the correct form:

    f(x) = \left\{<br />
     \begin{array}{lr}<br />
         x^2 & : x < 0\\<br />
       2x + 1 & : 0 \leq x \leq 2<br />
     \end{array}<br />
   \right.<br />
    Im stumped on proving continuity on the interval [1,2]. I can prove continuity at a point but dont know how to do it on a whole interval

    Heres what I have so far:
    I want to show if x_{0}\in [1,2], then \forall \epsilon >0, \exists \delta >0 such that |f(x)-f(x_{0})|<\epsilon provided that |x-x_{0}|<\delta and x\in[1,2]
    Actually Im not sure if x must exist in [1,2], any comment?
    Last edited by binkypoo; October 25th 2009 at 10:51 PM.
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  7. #7
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    So I let \delta=\frac{\epsilon}{2} to show f is continuous on [1,2] But I feel like delta should somehow involve the interval [1,2]... not sure where to go. any suggestions?
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