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Math Help - Limit with cubic factor

  1. #1
    Member jacs's Avatar
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    Limit with cubic factor

    Find \lim_{x\rightarrow0}\frac{(2+x)^{\frac13}-2^{\frac13}}{x}
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  2. #2
    Member jacs's Avatar
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    Limit with cubic factor

    Find  \lim_{x\to0}\frac{(2+x)^{\frac13}-2^{\frac13}}{x}

    I suspect somehow i need to utlise a^3-b^3=(a-b)(a^2+ab+b^2)

    although i am not sure how to do it

    thanks
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Now, a=2+x,\;b=2 .
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  4. #4
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    lim_{x\to 0}\frac{{(2+x)^{\frac{1}{3}}}-2^{\frac{1}{3}}}{(2+x)-2}

    which is one of generalised form

    lim_{x\to 0}\frac{x^n-a^n}{x-a} = na^{n-1}
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  5. #5
    Super Member TheChaz's Avatar
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    As has been noted above, this is the "alternate" definition of the derivative (or "derivative at a point")..
    If f(t) = t^(1/3)
    f'(t) is the above limit, with t in place of 2.
    So the above limit is f'(2) = (1/3)*2^(1/3-1)

    OR.
    View the numerator as one factor in the difference of cubes, and multiply by ...(something I'd rather not type) to "rationalize".
    Then you can use direct substitution.
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by jacs View Post
    Find  \lim_{x\to0}\frac{(2+x)^{\frac13}-2^{\frac13}}{x}

    I suspect somehow i need to utlise a^3-b^3=(a-b)(a^2+ab+b^2)

    although i am not sure how to do it

    thanks
    This may be way off bases but I have never seen limits taught in a pre-algebra glass!

    If you know about the derivative this this limit is

    \displaystyle \lim_{x \to 0}\frac{(2+x)^{\frac{1}{3}}-2^{\frac{1}{3}}}{x}=f'(2)

    Where f(x)=x^{\frac{1}{3}}.

    So as I said if you know about derivatives just take the derviative of f and plug in 2!
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