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Math Help - indeterminate

  1. #1
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    indeterminate

    Use analytical methods to evaluate the following problem.

    \lim_{x\rightarrow 4} \frac{^5\sqrt{6x+8}-2}{\frac{1}{x}+\frac{1}{4}}

    the answer to the limit is -\frac{6}{5}

    I have been taking a stab at it for quite a while. I think I figured it out but I a want to make sure the methods to how I got the answer were proper.

    First thing, Apply L' Hoptial's rule to make the -2 and the + \frac{1}{4} disappear

    \frac{(6x+8)^{1/5}-2}{x^{-1}+\frac{1}{4}} = \frac{\frac{1}{5}(6x+8)^{-4/5}*6}{-x^{-2}} =  \frac{\frac{6}{5}(6x+8)^{-4/5}}{-x^{-2}}


    I make powers have the same denominator
     -\frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{-4/5}}{x^{-\frac{10}{5}}} ]

    I get rid of the exponent denominator by applying an exponent of the power 5 to the limit. I also make it positive for simplicity.

     -\frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{-4/5}}{x^{-\frac{10}{5}}} ]^-5 =  -\frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{4}}{x^{10}} ]

    I observe the limit which will equal 1 and multiply by -6/5

     -\frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{4}}{x^{10}} ] = -\frac{6}{5} *1

    = -\frac{6}{5}
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: indeterminate

    That looks fine!

    Notice the denominator of the limit it has to be \frac{1}{x}-\frac{1}{4} else you don't have the indeterminate form \frac{0}{0} but offcourse it doesn't make a difference to apply l'Hopital's rule.
    Last edited by Siron; November 13th 2011 at 01:01 PM.
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