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Math Help - Another limits by change of variable problem

  1. #1
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    Another limits by change of variable problem

    I have two limits, to be solved by changing variables, that I'm having difficulty with:

    lim x-> 1

    x^(1/6) - 1
    _________

    x-1

    I let x^(1/6) = u. When x -> 1, u -> 1, and made the denominator u^6 - 1. However, this still gives a denominator of 0, so I'm not sure if I subbed the values correctly, or if I'm just not seeing how this should be factored. The other one, which I'm unsure how to do, is:

    lim x-> 1

    x^(1/6) - 1
    ________

    x^(1/3) - 1
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    Quote Originally Posted by starswept View Post
    I have two limits, to be solved by changing variables, that I'm having difficulty with:

    lim x-> 1

    x^(1/6) - 1
    _________

    x-1
    \lim_{x\to 1} \frac{x^{1/6} - 1}{x-1} = \lim_{y\to 1}\frac{y-1}{y^6-1} = \lim_{y\to 1} \frac{y-1}{(y-1)(y^5+y^4+y^3+y^2+y+1)}
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    Quote Originally Posted by starswept View Post
    I have two limits, to be solved by changing variables, that I'm having difficulty with:

    lim x-> 1

    x^(1/6) - 1
    _________

    x-1

    I let x^(1/6) = u. When x -> 1, u -> 1, and made the denominator u^6 - 1. However, this still gives a denominator of 0, so I'm not sure if I subbed the values correctly, or if I'm just not seeing how this should be factored.
    \lim_{x \to 1} \frac{x^{1/6} - 1}{x - 1}

    Let u = x^{1/6}

    Then we have
    \lim_{u \to 1} \frac{u - 1}{u^6 - 1}

    = \lim_{u \to 1} \frac{u - 1}{(u - 1)(u^5 + u^4 + u^3 + u^2 + u + 1)}

    = \lim_{u \to 1} \frac{1}{u^5 + u^4 + u^3 + u^2 + u + 1} = \frac{1}{6}

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by starswept View Post
    The other one, which I'm unsure how to do, is:

    lim x-> 1

    x^(1/6) - 1
    ________

    x^(1/3) - 1
    \lim_{x \to 1} \frac{x^{1/6} - 1}{x^{1/3} - 1}

    Let u = x^{1/6}

    Then you have
    \lim_{u \to 1} \frac{u - 1}{u^2 - 1}

    You should be able to finish this from here.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    \lim_{x \to 1} \frac{x^{1/6} - 1}{x - 1}

    Let u = x^{1/6}

    Then we have
    \lim_{u \to 1} \frac{u - 1}{u^6 - 1}

    = \lim_{u \to 1} \frac{u - 1}{(u - 1)(u^5 + u^4 + u^3 + u^2 + u + 1)}

    = \lim_{u \to 1} \frac{1}{u^5 + u^4 + u^3 + u^2 + u + 1} = \frac{1}{6}

    -Dan
    Or, you know, there's always L'Hopitals rule. (I'm going to duck all the rotten tomatos being thrown at me!)

    \lim_{x \to 1} \frac{x^{1/6} - 1}{x - 1} = \lim_{x \to 1} \frac{\frac{1}{6}x^{-5/6}}{1} = \frac{1}{6}

    all in one line!

    -Dan
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  6. #6
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    Quote Originally Posted by topsquark View Post
    Or, you know, there's always L'Hopitals rule. (I'm going to duck all the rotten tomatos being thrown at me!)

    \lim_{x \to 1} \frac{x^{1/6} - 1}{x - 1} = \lim_{x \to 1} \frac{\frac{1}{6}x^{-5/6}}{1} = \frac{1}{6}

    all in one line!

    -Dan
    That's a nifty one! Never seen it before Thanks!
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by starswept View Post
    That's a nifty one! Never seen it before Thanks!
    If you haven't yet, you will...

    -Dan
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