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Math Help - Limits

  1. #1
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    Limits

    So unfortunately, I missed one class and that was when we were learning about limits. The only thing I understand is that the limit does not exist if going towards an x value, the y value's are positive and negative.

    So I understand that this:

    would equal 1, right? (I hope, or I really need to get help with this stuff)

    But I don't understand what I'm supposed to do with this:

    Would it equal 2? it says x doesn't equal two... but does. I'm confused.

    Thanks to all who reply
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by forsheezy View Post
    So unfortunately, I missed one class and that was when we were learning about limits. The only thing I understand is that the limit does not exist if going towards an x value, the y value's are positive and negative.

    So I understand that this:

    would equal 1, right? (I hope, or I really need to get help with this stuff)

    But I don't understand what I'm supposed to do with this:

    Would it equal 2? it says x doesn't equal two... but does. I'm confused.

    Thanks to all who reply
    For the first one direct substitution of the limit yields a determinant form. Determinant form just means it has no problems like divide by zero etc.

    So \lim_{x\to{1}}x+2=(1)+2=3


    For the second one a limit is defined as the value as x\to{c} what f(x)\to

    So the exact value at x=2 is not equivalent to limit since f(x) is not continous at x=2

    But \lim_{x\to{c}}f(x)\text{ }\exists\text{ }\text{iff}\text{ }\lim_{x\to{c^{-}}}f(x)=\lim_{x\to{c^{+}}}f(x)

    And as it does \lim_{x\to{2^{-}}}f(x)=4-(2)=2=\lim_{x\to{2^{+}}}f(x)

    \therefore\lim_{x\to{2}}f(x)=2
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  3. #3
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    May 2008
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    Thank You
    I somewhat understand it now!
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