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Math Help - Hard Limit

  1. #1
    Senior Member polymerase's Avatar
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    Hard Limit

    How do get the following

    \displaystyle\lim_{x\to 1}\frac{x^\frac{1}{7}-1}{x^\frac{1}{5}-1}

    Without l'Hopital's Rule....
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  2. #2
    Eater of Worlds
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    One way to appraoch it is to do some fancy factoring.

    \lim_{x\rightarrow{1}}\frac{x^{\frac{1}{7}}-1}{x^{\frac{1}{5}}-1}

    = \lim_{x\rightarrow{1}}\frac{x^{\frac{4}{35}}+x^{\f  rac{3}{35}}+x^{\frac{2}{35}}+x^{\frac{1}{35}}+1}{x  ^{\frac{6}{35}}+x^{\frac{4}{35}}+x^{\frac{3}{35}}+  x^{\frac{2}{35}}+x^{\frac{1}{7}}+x^{\frac{1}{35}}+  1}=\boxed{\frac{5}{7}}
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  3. #3
    Math Engineering Student
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    Quote Originally Posted by polymerase View Post
    \displaystyle\lim_{x\to 1}\frac{x^\frac{1}{7}-1}{x^\frac{1}{5}-1}
    Substitute u^{35}=x, the limit becomes to

    \lim_{u\to1}\frac{u^5-1}{u^7-1}.

    The factor u-1 is bothering the top & bottom, so we pull it out:

    u^5-1=(u-1)(u^4+u^3+u^2+u+1), u^7-1=(u-1)(u^6+u^5+u^4+u^3+u^2+u+1), so

    \lim_{u\to1}\frac{u^5-1}{u^7-1}=\lim_{u\to1}\frac{u^4+u^3+u^2+u+1}{u^6+u^5+u^4+  u^3+u^2+u+1},

    and the conclusion follows \blacksquare
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  4. #4
    Senior Member polymerase's Avatar
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    Quote Originally Posted by galactus View Post
    One way to appraoch it is to do some fancy factoring.

    \lim_{x\rightarrow{1}}\frac{x^{\frac{1}{7}}-1}{x^{\frac{1}{5}}-1}

    = \lim_{x\rightarrow{1}}\frac{x^{\frac{4}{35}}+x^{\f  rac{3}{35}}+x^{\frac{2}{35}}+x^{\frac{1}{35}}+1}{x  ^{\frac{6}{35}}+x^{\frac{4}{35}}+x^{\frac{3}{35}}+  x^{\frac{2}{35}}+x^{\frac{1}{7}}+x^{\frac{1}{35}}+  1}=\boxed{\frac{5}{7}}
    How do you do that....what's the "rule"(lack of a better word)?
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  5. #5
    Eater of Worlds
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    It's basically the same as the previous post, but K eliminated the fractional exponents. Just factoring. See the pattern?.
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  6. #6
    Senior Member polymerase's Avatar
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    Quote Originally Posted by galactus View Post
    It's basically the same as the previous post, but K eliminated the fractional exponents. Just factoring. See the pattern?.
    I get K's completely but i dont see the pattern for yours
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by polymerase View Post
    I get K's completely but i dont see the pattern for yours
    Its just writting x^{1/7}-1 and x^{1/5}-1 as sums of powers of x^{1/35} , which can be done because 5\times 7=35

    RonL
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