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Math Help - difficult limit

  1. #1
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    difficult limit

    How can I solve the folowing limit using only algebric manipulation:
    limit when x aproches 0 of: {[x^(1/3)]-1}/{x^(1/2)]-1}
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  2. #2
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    Quote Originally Posted by Mppl View Post
    How can I solve the folowing limit using only algebric manipulation:
    limit when x aproches 0 of: {[x^(1/3)]-1}/{x^(1/2)]-1}
    Dear Mppl,

    Do you mean, \displaystyle\lim_{x\rightarrow{0}}\frac{x^{\frac{  1}{3}}-1}{x^{\frac{1}{2}}-1} ??
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  3. #3
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    yes precisely that one. Can you help me with that?
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  4. #4
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    Question: Why do you say this is difficult?
    Do you realize that  \displaystyle \lim _{x \to 0} \sqrt x does not exist because \sqrt{x} is not defined for x<0.
    But  \displaystyle \lim _{x \to 0^+} \sqrt x=0 .

    So you may want to double check the statement of this question.
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  5. #5
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    well lets admit its 0+ then...whats the result then? and how to get there?
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  6. #6
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    Quote Originally Posted by Mppl View Post
    well lets admit its 0+ then...whats the result then? and how to get there?
     \displaystyle \lim _{x \to 0^+} \sqrt x=0 .
     \displaystyle \lim _{x \to 0^+} \sqrt[3] x=0 .

    Now you finish.
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  7. #7
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    oh my bad the limit is x aproching 1 and not 0 sorry...what would be the solution there?
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  8. #8
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    Let x = y^6 so then you have

    \displaystyle \lim_{y \to 1} \dfrac{y^2-1}{y^3-1}

    which you can factor but L'Hopital's rule would be faster.
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  9. #9
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    Let's make the problem a bit more interesting and assume that the limit is supposed to be as x goes to 1 rather than 0. You can then check using l'Hpital's rule that the answer should be 2/3. But it has to be done using only algebraic manipulation, so l'Hp is not allowed.

    Let u = x^{1/6}. Then as x goes to 1, u also goes to 1. So we want to find

    \displaystyle \lim_{x\to 1}\frac{x^{1/3}-1}{x^{1/2}-1} = \lim_{u\to 1}\frac{u^2-1}{u^3-1} = \lim_{u\to 1}\frac{(u-1)(u+1)}{(u-1)(u^2+u+1)} = \lim_{u\to 1}\frac{u+1}{u^2+u+1} = \frac23.

    Edit. Beaten by Danny.
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  10. #10
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    thanks a lot men
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