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    I need show that f(x)= x^2 +2 is continuous at x=3.
    So I'm trying to prove that as x approaches 3 from the left = the value as x approaches 3 from the right.
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    The other one is upside down
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    (Original post by Vorsah)
    The other one is upside down
    Obviously that approach will work if the left and right hand limits are equal to each other AND equal to the value of the function at that point. However, that's certainly a more complicated way of doing it.

    The definition of continuity states that

    \displaystyle \lim_{x \rightarrow a} f(x) = f(a)

    and the epsilon-delta version is

    \forall \epsilon > 0: \exists \delta:  |x-a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon

    So this is what you should be after

    |x-3| < \delta \Rightarrow |x^2 + 2 - 11| = |x^2 - 9| < \epsilon

    It should be easy for you to bound that by using the difference of two squares. That will give you delta in terms of epsilon.
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    (Original post by Indeterminate)
    That's not the way to prove continuity. You can have functions whose left and right hand limits agree at points where the "ordinary" limit is something completely different (i.e there's a discontinuity).

    The definition of continuity states that

    \displaystyle \lim_{x \rightarrow a} f(x) = f(a)
    But that's equivalent to \displaystyle \lim_{x \rightarrow a+} f(x) = f(a) and \displaystyle \lim_{x \rightarrow a-} f(x) = f(a) which is what he is trying to prove, I think.
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    (Original post by atsruser)
    But that's equivalent to \displaystyle \lim_{x \rightarrow a+} f(x) = f(a) and \displaystyle \lim_{x \rightarrow a-} f(x) = f(a) which is what he is trying to prove, I think.
    That's a possibility, though you'd usually expect someone to use the normal definition in order to make life easier for themselves.

    EDIT: I've edited my post above to incorporate this.
 
 
 
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