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Math Help - Limit of a function

  1. #1
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    Limit of a function

    Hi
    Could someone please help me with the general method of finding limits like this one:

    x-> infinity of x^(1/3) [ (x+8)^(2/3) - (x+7)^(2/3)]
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  2. #2
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     \lim_{n\rightarrow\infty} {(x+8)}^{2/3} - {(x+7)}^{2/3} will give you  \infty - \infty , which is an indeterminate of L'H˘pital's rule. You have to algebraically change it up so as  \lim_{n\rightarrow\infty} \left({(x+8)}^{2/3} - {(x+7)}^{2/3}\right) you get  \frac{0}{0} or  \frac{\infty}{\infty} . Finally, take the derivative of the function.
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  3. #3
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    Quote Originally Posted by appletree View Post
    Hi
    Could someone please help me with the general method of finding limits like this one:

    x-> infinity of x^(1/3) [ (x+8)^(2/3) - (x+7)^(2/3)]


    Take note that for x>0\,,\,\,x^{4/3}(A+B)^{4/3}=(Ax+Bx)^{4/3} , and also that A-B=\left(A^{1/3}-B^{1/3}\right)\left(A^{2/3}+A^{1/3}B^{1/3}+B^{1/3}\right) :

    x^{1/3} [ (x+8)^{2/3} - (x+7)^{2/3}]= x^{1/3}\,\frac{(x+8)^2-(x-7)^2}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}= x^{1/3}\,\frac{2x+15}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}=

    \frac{2x^{4/3}+15x^{1/3}}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}\,\frac{1/x^{4/3}}{1/x^{4/3}}= \frac{2+15/x}{\left(1+8/x\right)^{4/3}+\left(1+8/x\right)^{2/3}\left(1+7/x\right)^{2/3}+\left(1+7/x\right)^{4/3}} \xrightarrow [x\to\infty]{}2/3

    Tonio
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