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Math Help - Limits

  1. #1
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    Exclamation Limits

    Help me solve this please:

    lim [(x+1)^2 - 1] / x
    x-> 0



    I got 0. I don't think that's right though.
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  2. #2
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    Re: Limits

    Quote Originally Posted by habibixox View Post
    Help me solve this please:

    lim [(x+1)^2 - 1] / x
    x-> 0



    I got 0. I don't think that's right though.
    Start by expanding the numerator and collecting like terms, then dividing each term by \displaystyle x, before trying to let \displaystyle x \to 0...
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  3. #3
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    Re: Limits

    (x+1)(x+1) - 1 / x

    what i did was:

    Lim lim [x+1] ^2 - lim 1 / lim x
    x-> 0 x->0 x->0 x-> 0
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    Re: Limits

    Quote Originally Posted by habibixox View Post
    (x+1)(x+1) - 1 / x

    what i did was:

    Lim lim [x+1] ^2 - lim 1 / lim x
    x-> 0 x->0 x->0 x-> 0
    Obviously what you have done is not working, since you are getting 0/0, an indeterminate form. Why not try following the advice you were given?
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    Re: Limits

    i got 2.
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    Re: Limits

    Quote Originally Posted by habibixox View Post
    i got 2.
    Would you mind showing how you got an answer of 2 please?
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  7. #7
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    Re: Limits

    (x+1) (x+1) =
    x^2+2x+1 -1 / x

    then
    x(x+2) / x

    then cancel the x's

    0 +2 = 2
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  8. #8
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    Re: Limits

    Quote Originally Posted by habibixox View Post
    (x+1) (x+1) =
    x^2+2x+1 -1 / x

    then
    x(x+2) / x

    then cancel the x's

    0 +2 = 2
    Assuming you meant (x + 1)(x + 1) + 1 = x^2 + 2x + 1 - 1 then you are correct Well done.
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  9. #9
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    Re: Limits

    thank you!
    so ultimately there are only three ways of solving limits

    by factoring
    substitution
    and conjugation
    right?

    and you only use conjugation when there is a square root right and a number?
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  10. #10
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    Re: Limits

    Quote Originally Posted by habibixox View Post
    thank you!
    so ultimately there are only three ways of solving limits

    by factoring
    substitution
    and conjugation
    right?

    and you only use conjugation when there is a square root right and a number?
    There are many more ways than three to solve limits. The easiest way is to find some way to simplify the function so that you can factor and cancel any terms that will end up making an indeterminate form, which like you said, might involve factoring or conjugation.
    Substitution is only used if the function is continuous at the point you are approaching, or once you have done enough to be able to substitute without resulting in an indeterminate form.
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