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Thread: plase help

  1. #1
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    plase help

    <1> Lim ln(2x-1) / sin (pi)x
    x->1

    <2>Lim [((x+1)/ln(x+1)) -1/x)]
    x->0
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bobby77
    <1> Lim ln(2x-1) / sin (pi)x
    x->1
    What is:

    $\displaystyle
    \lim_{x \rightarrow 1} \frac{\ln(2x-1)}{\sin(\pi x)}?
    $

    As:

    $\displaystyle
    \lim_{x \rightarrow 1} \ln(2x-1)=\ln(1)=0
    $,

    and

    $\displaystyle
    \lim_{x \rightarrow 1} \sin(\pi x)=0
    $

    we need to use L'Hopital's rule to evaluate the limit so we need:

    $\displaystyle
    \lim_{x \rightarrow 1} \frac{d}{dx}\ln(2x-1)=\lim_{x \rightarrow 1} \frac{2}{2x-1}=2
    $,

    and:

    $\displaystyle
    \lim_{x \rightarrow 1} \frac{d}{dx}\sin(\pi x)=\lim_{x \rightarrow 1}\pi \cos(\pi x)=- \pi
    $.

    Then L'Hopital's rule tells us that:

    $\displaystyle
    \lim_{x \rightarrow 1} \frac{\ln(2x-1)}{\sin(\pi x)}=\frac{\lim_{x \rightarrow 1} \frac{d}{dx}\ln(2x-1)}{\lim_{x \rightarrow 1} \frac{d}{dx}\sin(\pi x)}=-\frac{2}{\pi}
    $.

    Finally checking with numerical experiments indicates that this is right.

    RonL
    Last edited by CaptainBlack; Mar 1st 2006 at 06:03 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by bobby77
    <2>Lim [((x+1)/ln(x+1)) -1/x)]
    x->0
    What is:

    $\displaystyle
    \lim_{x \rightarrow 0} \left\{ \frac{x+1}{\ln(x+1)}-\frac{1}{x} \right\}?
    $

    This is indeterminate as it is so rearrange:

    $\displaystyle
    \lim_{x \rightarrow 0} \left\{ \frac{x(x+1)-\ln(x+1)}{x \ln(x+1)} \right\}
    $.

    Now apply L'Hopital's rule:

    $\displaystyle
    \lim_{x \rightarrow 0} \left\{ \frac{x(x+1)-\ln(x+1)}{x \ln(x+1)} \right\}=$$\displaystyle
    \frac{\lim_{x \rightarrow 0} \frac{d}{dx}\{x(x+1)-\ln(x+1)\}}{\lim_{x \rightarrow 0} \frac{d}{dx} \{x \ln(x+1)\} }=\frac{0}{0}
    $

    which is indeterminate, so we need the second derivatives:

    $\displaystyle
    \lim_{x \rightarrow 0} \left\{ \frac{x(x+1)-\ln(x+1)}{x \ln(x+1)} \right\}=$$\displaystyle
    \frac{\lim_{x \rightarrow 0} \frac{d^2}{dx^2}\{x(x+1)-\ln(x+1)\}}{\lim_{x \rightarrow 0} \frac{d^2}{dx^2} \{x \ln(x+1)\} }=\frac{3}{1}=1.5
    $

    Which again checks out numericaly

    RonL
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