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Math Help - Limit using L'Hospitals Rule

  1. #1
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    Limit using L'Hospitals Rule

    lim (1+C/x)^(hx)
    x-> infinity

    I have to evaluate a limit

    since if you plug in infinity/infinity you have to use L'Hospitals rule

    Thank you
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Zvaigzdute View Post
    lim (1+C/x)^(hx)
    x-> infinity

    I have to evaluate a limit

    since if you plug in infinity/infinity you have to use L'Hospitals rule

    Thank you
    \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{c}{x}} \right)^{hx}} = \mathop {\lim }\limits_{x \to \infty } \exp \ln {\left( {1 + \frac{c}{x}} \right)^{hx}} = \mathop {\lim }\limits_{x \to \infty } \exp \left\{ {hx\ln \left( {1 + \frac{c}{x}} \right)} \right\} =

    = \mathop {\lim }\limits_{x \to \infty } \exp \frac{{\ln \left( {1 + \frac{c}{x}} \right)}}{{\frac{1}{{hx}}}} = \mathop {\lim }\limits_{x \to \infty } \exp \frac{{\frac{d}{{dx}}\ln \left( {1 + \frac{c}{x}} \right)}}{{\frac{d}{{dx}}\frac{1}{{hx}}}} = \mathop {\lim }\limits_{x \to \infty } \exp \frac{ch}{{1 + \frac{c}{x}}} =e^{ch}.
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  3. #3
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    Hello, Zvaigzdute!

    Another approach . . .


    Evaluate: . \lim_{x\to\infty}\left(1+\frac{C}{x}\right)^{hx}
    Let: y \;=\;\left(1 + \frac{C}{x}\right)^{hx}

    Take logs: . \ln y \;=\;\ln\left(1 + \frac{C}{x}\right)^{hx} \;=\;hx\cdot\ln\left(1 + \frac{C}{x}\right) \;= . \frac{h\ln\left(1 + \frac{C}{x}\right)}{\frac{1}{x}} \;=\;\frac{h\ln\left(1 + Cx^{-1}\right)}{x^{-1}}

    Apply l'Hopital: . \frac{h\cdot\dfrac{1}{1+Cx^{-1}}\cdot(-Cx^{-2})} {-x^{-2}} \;=\;\frac{Ch}{1 + \frac{C}{x}}

    Then: . \lim_{x\to\infty}\ln y \;=\;\lim_{x\to\infty} \left(\frac{Ch}{1 + \frac{C}{x}}\right) \;=\;\frac{Ch}{1+0} \;=\;Ch


    Therefore: . \ln y \:=\:Ch \quad\Rightarrow\quad y \;=\;e^{Ch}

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