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Math Help - limit question

  1. #1
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    limit question

    <br />
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}<br />

    i get infinity - infinity
    its undefined
    ??
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    <br />
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}<br />

    i get infinity - infinity
    its undefined
    ??
    You need to get this into the form of a fraction to apply L'Hopitals rule. I think multiplying this by its conjugate divided by itself should work. It should come out to \frac{(x+2)^{\frac{4}{3}}-(x-2)^{\frac{4}{3}}}{(x+2)^{\frac{2}{3}}+(x-2)^{\frac{2}{3}}}
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  3. #3
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    Quote Originally Posted by transgalactic View Post
    <br />
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}<br />

    i get infinity - infinity
    its undefined
    ??
    It's not undefined. It's an indeterminate form.

    The limit is 0 (and I'm sure you will recognise this answer ....)

    One approach might be to simply apply the generalised binomial theorem to each term and simplify. Then take the limit ....


    Another approach might be to re-write as x^{2/3} \left[ \left( 1 + \frac{2}{x}\right)^{2/3} - \left( 1 - \frac{2}{x}\right)^{2/3}\right] = \frac{\left( 1 + \frac{2}{x}\right)^{2/3} - \left( 1 - \frac{2}{x}\right)^{2/3}}{x^{-2/3}} so that you have the indeterminant form  \frac{0}{0} and then use l'Hopital's Rule.


    Quote Originally Posted by Jameson View Post
    You need to get this into the form of a fraction to apply L'Hopitals rule. I think multiplying this by its conjugate divided by itself should work. It should come out to \frac{(x+2)^{\frac{4}{3}}-(x-2)^{\frac{4}{3}}}{(x+2)^{\frac{2}{3}}+(x-2)^{\frac{2}{3}}}
    I'm not so sure this works. The numerator has the form \infty - \infty and so you have the same troubles but worse algebra ....
    Last edited by mr fantastic; February 27th 2009 at 04:43 AM.
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  4. #4
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    Write \underset{x\to \infty }{\mathop{\lim }}\,\big((x+2)^{2/3}+x^{2/3}-(x-2)^{2/3}-x^{2/3}\big), then consider \sqrt[3]{(x+2)^{2}}-\sqrt[3]{x^{2}} and use the identity a^{3}-b^{3}=(a-b)\left( a^{2}+ab+b^{2} \right). The rest is algebra.
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