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Thread: Limit: Indeterminate Form/L'Hospital's Rule

  1. #1
    Member RedBarchetta's Avatar
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    Limit: Indeterminate Form/L'Hospital's Rule

    Evaluate the limit. (If you need to use - or , enter -INFINITY or INFINITY.)

    $\displaystyle
    \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {\ln \left( {4x} \right)} \right)}}
    {{4x}}
    $

    Here's what I tried. Here's our first rule:

    $\displaystyle
    \begin{gathered}
    \mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}
    {{g(x)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{f'(x)}}
    {{g'(x)}} \hfill \\
    \mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
    {{\ln 4x}} \cdot \tfrac{1}
    {{4x}} \cdot \tfrac{4}
    {1}}}
    {4} \hfill \\
    \mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
    {{x\ln 4x}}}}
    {4} \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    \mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
    {{x\ln 4x}}}}
    {4} \hfill \\
    \mathop {\lim }\limits_{x \to \infty } \frac{1}
    {{4x\ln 4x}} \hfill \\
    \end{gathered}
    $

    So then what? It's not infinity or -infinity.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
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    5
    Quote Originally Posted by RedBarchetta View Post
    Evaluate the limit. (If you need to use - or , enter -INFINITY or INFINITY.)

    $\displaystyle
    \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {\ln \left( {4x} \right)} \right)}}
    {{4x}}
    $

    Here's what I tried. Here's our first rule:

    $\displaystyle
    \begin{gathered}
    \mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}}
    {{g(x)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{f'(x)}}
    {{g'(x)}} \hfill \\
    \mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
    {{\ln 4x}} \cdot \tfrac{1}
    {{4x}} \cdot \tfrac{4}
    {1}}}
    {4} \hfill \\
    \mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
    {{x\ln 4x}}}}
    {4} \hfill \\
    \end{gathered}
    $

    $\displaystyle
    \begin{gathered}
    \mathop {\lim }\limits_{x \to \infty } \frac{{\tfrac{1}
    {{x\ln 4x}}}}
    {4} \hfill \\
    \mathop {\lim }\limits_{x \to \infty } \frac{1}
    {{4x\ln 4x}} \hfill \\
    \end{gathered}
    $

    So then what? It's not infinity or -infinity.
    $\displaystyle \lim_{x\to{\infty}}\frac{\ln(\ln 4x)}{4x}$

    $\displaystyle \lim_{x\to{\infty}}\frac{\frac{1}{x\ln 4x}}{4}$

    $\displaystyle \lim_{x\to{\infty}}\frac{1}{4x\ln 4x}=\color{red}\boxed{0}$. This the case because the denominator approaches infinity : $\displaystyle \frac{1}{\infty}\rightarrow 0$.
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