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Math Help - "very difficult" limits question

  1. #1
    Senior Member polymerase's Avatar
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    "very difficult" limits question

    <br />
\displaystyle\lim_{x\to\infty}\sqrt[3]{(x^3+5x^2)}-x<br />
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  2. #2
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    How about multiplyin' by

    \frac{{\sqrt[3]{{\left( {x^3  + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3  + 5x^2 }} + x^2 }}<br />
{{\sqrt[3]{{\left( {x^3  + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3  + 5x^2 }} + x^2 }}?
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    How about multiplyin' by

    \frac{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}<br />
{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}?
    huh?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    huh?
    He's showing you how to rationalize the "numerator" of your expression. (It requires more terms for a cube root than it does for a square root.) Try it and see what happens.

    -Dan
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  5. #5
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    Quote Originally Posted by polymerase View Post
    <br />
\displaystyle\lim_{x\to\infty}\sqrt[3]{(x^3+5x^2)}-x<br />
    If you know these methods you can try:

    <br />
\sqrt[3]{x^3+5x^2}-x=x~ \sqrt[3] {1+5/x} -x<br />

    Expand the cube root using the binomial theorem:

    <br />
\sqrt[3]{x^3+5x^2}-x=x~ \sqrt[3] {1+5/x} -x=x~\left(1+\frac{5}{3x}+O(x^{-2})\right) -x=\frac{5}{3}+O(x^{-1})<br />

    Hence the required limit is 5/3.

    RonL
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  6. #6
    Senior Member polymerase's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    If you know these methods you can try:

    <br />
\sqrt[3]{x^3+5x^2}-x=x~ \sqrt[3] {1+5/x} -x<br />

    Expand the cube root using the binomial theorem:

    <br />
\sqrt[3]{x^3+5x^2}-x=x~ \sqrt[3] {1+5/x} -x=x~\left(1+\frac{5}{3x}+O(x^{-2})\right) -x=\frac{5}{3}+O(x^{-1})<br />

    Hence the required limit is 5/3.

    RonL
    how did u use the binomial theorem when its a fractional exponent?
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  7. #7
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    How about multiplyin' by

    \frac{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}<br />
{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}?
    I tried this way and in the end, I end up with a nice numerator but not so nice denominator:

    \dfrac{5x^2}{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }

    So how would u deal with the denominator now?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    I tried this way and in the end, I end up with a nice numerator but not so nice denominator:

    \dfrac{5x^2}{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }

    So how would u deal with the denominator now?
    Multiply the numerator and denominator by \frac{1}{x^2}, then take your limit.

    -Dan
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  9. #9
    Senior Member polymerase's Avatar
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    Quote Originally Posted by topsquark View Post
    Multiply the numerator and denominator by \frac{1}{x^2}, then take your limit.

    -Dan
    thanks
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by polymerase View Post
    how did u use the binomial theorem when its a fractional exponent?
    The binomial theorem applies for any real exponent.

    A MacLaurin/Taylor expansion will do just as well if you don't like a binomial expansion

    RonL
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  11. #11
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    How about multiplyin' by

    \frac{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}<br />
{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}?
    Your method gives u this:

    <br />
\frac{5}{1+\(1+\frac{5}{x}}<br />
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  12. #12
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    How about multiplyin' by

    \frac{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}<br />
{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + \sqrt[3]{{x^3 + 5x^2 }} + x^2 }}?
    Using your method gives me this:

    <br />
\lim_{x\to\infty}\dfrac{5}{\sqrt[3]{{\left( {1 + \frac{5}{x} } \right)^2 }} + \sqrt[3]{{\frac{1}{x^3} + \frac{5}{x^4} }} + 1 }<br />
    and when you take the limit to infinity you get \frac{5}{2} but the answer is \frac{5}{3}
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  13. #13
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    You're wrong
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  14. #14
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    Using your method gives me this:

    <br />
\lim_{x\to\infty}\dfrac{5}{\sqrt[3]{{\left( {1 + \frac{5}{x} } \right)^2 }} + \sqrt[3]{{\frac{1}{x^3} + \frac{5}{x^4} }} + 1 }<br />
    and when you take the limit to infinity you get \frac{5}{2} but the answer is \frac{5}{3}
    Check that again:
    \lim_{x \to \infty} \dfrac{5x^2}{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }

    = \lim_{x \to \infty} \dfrac{5}{\frac{1}{x^2}\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + \frac{1}{x}\sqrt[3]{{x^3 + 5x^2 }} + 1 }

    = \lim_{x \to \infty} \dfrac{5}{\sqrt[3]{{\left( {1 + \frac{5}{x} } \right)^2 }} + \sqrt[3]{{1 + \frac{5}{x} }} + 1 }

    = \frac{5}{1 + 1 + 1} = \frac{5}{3}

    -Dan
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  15. #15
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    You're wrong
    ok ok your right....my bad i forgot the x in the 2 term
    Thanks man
    ....wow how do you come up with something like this to solve this question? i have never seen anyone rationalize like this before
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