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Thread: l'Hopital's rule....help

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

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

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

    So then

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

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

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

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

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

    So then

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

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

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

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