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Math Help - l'Hopital's rule....help

  1. #1
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    Post l'Hopital's rule....help

    Find the lim as 'x' approaches infinity (1+2x)^{(17)/(2lnx)} (sorry dont know how to put the lim on here....) thanks for any help..
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by omibayne View Post
    Find the lim as 'x' approaches infinity (1+2x)^{(17)/(2lnx)} (sorry dont know how to put the lim on here....) thanks for any help..
    You have \lim_{x\to\infty}\left(1+2x\right)^{\frac{17}{2\ln  (x)}}

    Must you use L'hopital's? To make it a little simpler try letting \ln(x)=z so our limit becomes \lim_{z\to\infty}\left(1+2e^z\right)^{\frac{17}{2z  }}=L\Rightarrow{L^{\frac{2}{17}}}=\lim_{z\to\infty  }\left(1+2e^z\right)^{\frac{1}{z}}

    So then

    \ln\left(L^{\frac{2}{17}}\right)=\lim_{z\to\infty}  \frac{\ln\left(1+2e^z\right)}{z}

    Either applying L'hopitals (or noticing that \ln\left(1+2e^z\right)\sim{z}) gives us

    \lim_{z\to\infty}\frac{\ln\left(1+2e^z\right)}{z}=  \lim_{z\to\infty}\frac{\frac{2e^z}{1+2e^z}}{1}=1

    So if \ln\left(L^{\frac{2}{17}}\right)=1\Rightarrow{L=e^  {\frac{17}{2}}}
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    You have \lim_{x\to\infty}\left(1+2x\right)^{\frac{17}{2\ln  (x)}}

    Must you use L'hopital's? To make it a little similar try letting \ln(x)=z so our limit becomes \lim_{z\to\infty}\left(1+2e^z\right)^{\frac{17}{2z  }}=L\Rightarrow{L^{\frac{2}{17}}}=\lim_{z\to\infty  }\left(1+2e^z\right)^{\frac{1}{z}}

    So then

    \ln\left(L^{\frac{2}{17}}\right)=\lim_{z\to\infty}  \frac{\ln\left(1+2e^z\right)}{z}

    Either applying L'hopitals (or noticing that \ln\left(1+2e^z\right)\sim{z}) gives us

    \lim_{z\to\infty}\frac{\ln\left(1+2e^z\right)}{z}=  \lim_{z\to\infty}\frac{\frac{2e^z}{1+2e^z}}{1}=1

    So if \ln\left(L^{\frac{2}{17}}\right)=1\Rightarrow{L=e^  {\frac{17}{2}}}
    thanks alot for ur help...
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