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Math Help - limit question

  1. #1
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    Post limit question

    Hey having trouble with this one guys.

    \dfrac{lim}{x \rightarrow -2} :              \dfrac{x}{x+2}

    Cheers
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  2. #2
    Super Member Anonymous1's Avatar
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    Definitely the wrong forum, but I ain't the po-lice.

    \frac{x}{x+2} = \frac{1}{1+2/x}

    Wait that did nothing..? One sec.

    Oh, approach it from the right and left and see what the deal is.

    BTW \lim_{x\rightarrow -2} = \lim_{x\rightarrow -2}
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by smplease View Post
    Hey having trouble with this one guys.

    \dfrac{lim}{x \rightarrow -2} :              \dfrac{x}{x+2}

    Cheers

    what about dividing by x to get

    \dfrac{lim}{x \rightarrow -2} :   \frac{1}{1+(\frac{2}{x})}
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  4. #4
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by harish21 View Post
    what about dividing by x to get

    \dfrac{lim}{x \rightarrow -2} :   \frac{1}{1+(\frac{2}{x})}
    What does this accomplish?
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Anonymous1 View Post
    What does this accomplish?


    \dfrac{lim}{x \rightarrow (-2^-)} :              \dfrac{x}{x+2} = \infty

    \dfrac{lim}{x \rightarrow (-2^+)} :              \dfrac{x}{x+2} = -\infty


    the limit does not exist!
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  6. #6
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by harish21 View Post
    \dfrac{lim}{x \rightarrow (-2^-)} :              \dfrac{x}{x+2} = \infty

    \dfrac{lim}{x \rightarrow (-2^+)} :              \dfrac{x}{x+2} = -\infty


    the limit does not exist!
    When this happens you average them.

    E[Limit] = \frac{\infty - \infty}{2} = 0. So the limit is 0!

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    Don't worry I'm not serious.
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  7. #7
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    Thanks guys.

    I thought I must have been doing it wrong because I kept getting undefined answers but this explains why
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