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Thread: Limit Problem - # 3

  1. #1
    MHF Contributor Jason76's Avatar
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    Limit Problem - # 3

    $\displaystyle \lim x \rightarrow 0^{+}[\sin x \ln 2x]$

    $\displaystyle \lim x \rightarrow 0^{+}[\sin (0) \ln 2(0)]$

    $\displaystyle \lim x \rightarrow 0^{+}[(0)( \ln 0)]$

    $\displaystyle \lim x \rightarrow 0^{+}[(0)( -\infty)]$ Indeterminate

    $\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\ln2 x}{\dfrac{1}{\sin x}}] $ Rewrite

    $\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\ln2 x}{\csc x}] $

    $\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{d}{dx} \ln x}{\dfrac{d}{dx} \csc x}]$

    $\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{1}{x}}{-\csc x \cot x}]$

    $\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{1}{0}}{-\csc 0 \cot 0}]$
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  2. #2
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    Re: Limit Problem - # 3

    Quote Originally Posted by Jason76 View Post
    $\displaystyle \lim x \rightarrow 0^{+}[\sin x \ln 2x]$

    $\displaystyle \lim x \rightarrow 0^{+}[\sin (0) \ln 2(0)]$

    $\displaystyle \lim x \rightarrow 0^{+}[(0)( \ln 0)]$

    $\displaystyle \lim x \rightarrow 0^{+}[(0)( -\infty)]$ Indeterminate

    $\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\ln2 x}{\dfrac{1}{\sin x}}] $ Rewrite

    $\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\ln2 x}{\csc x}] $

    $\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{d}{dx} \ln x}{\dfrac{d}{dx} \csc x}]$

    $\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{1}{x}}{-\csc x \cot x}]$

    $\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{1}{0}}{-\csc 0 \cot 0}]$
    $\displaystyle \displaystyle \begin{align*} \sin{(x)}\ln{(2x)} &= \frac{\sin{(x)}}{\left[ \ln{(2x)} \right] ^{-1} } \end{align*}$

    This is now of the form $\displaystyle \displaystyle \begin{align*} \frac{0}{0} \end{align*}$ so L'Hospital's Rule can be applied.
    Thanks from topsquark
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