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Math Help - Limits Q

  1. #1
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    Limits Q

    Use the inequality 0< \int_{1}^{\sqrt{x}}\frac{dt}{t}< \sqrt{x} to show \lim_{x\rightarrow \infty }(\frac{\ln x}{x})=0
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by vuze88 View Post
    Use the inequality 0< \int_{1}^{\sqrt{x}}\frac{dt}{t}< \sqrt{x} to show \lim_{x\rightarrow \infty }(\frac{\ln x}{x})=0
    0 < \int_1^{\sqrt x } {\frac{{dt}}{t}}  < \sqrt x  \,\Leftrightarrow \, 0 < \ln \sqrt x  < \sqrt x \, \Leftrightarrow

    \Leftrightarrow 0 < \ln x < 2\sqrt x  \Leftrightarrow 0 < \frac{{\ln x}}<br />
{x} < \frac{2}<br />
{{\sqrt x }} \Leftrightarrow

    \Leftrightarrow 0 < \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}<br />
{x} < \mathop {\lim }\limits_{x \to \infty } \frac{2}<br />
{{\sqrt x }} = 0 \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{\ln x}}<br />
{x} = 0.
    Last edited by DeMath; October 2nd 2009 at 08:31 PM. Reason: typos
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by vuze88 View Post
    Use the inequality 0< \int_{1}^{\sqrt{x}}\frac{dt}{t}< \sqrt{x} to show \lim_{x\rightarrow \infty }(\frac{\ln x}{x})=0
    Evaluating the integral gives us 0<\ln(\sqrt{x})<\sqrt{x}\implies 0<\frac{1}{2}\ln(x)<\sqrt{x}.

    Note that \lim_{x\to\infty}\frac{\ln(x)}{x} = 2\lim_{x\to\infty}\frac{\frac{1}{2}\ln(x)}{x}

    The above equations imply that

    0<\lim_{x\to\infty}\frac{\ln(x)}{x}<2\lim_{x\to\in  fty}\frac{\sqrt{x}}{x}

    Now we know that 2\lim_{x\to\infty}\frac{\sqrt{x}}{x}=2\lim_{x\to\i  nfty}\frac{1}{\sqrt{x}}=0, so we have successfully squeezed the limit:

    0<\lim_{x\to\infty}\frac{\ln(x)}{x}<0

    So it equals 0. \square
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