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Thread: Limit Problem

  1. #1
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    Limit Problem

    Calculate the following limit

    $\displaystyle \lim_{x \to 0} \ (x^2) cosh (1/x) $
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  2. #2
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    when x tends to 0 , $\displaystyle x^2 \cosh(\frac{1}{x}) $ tends to infinity .

    Since $\displaystyle \cosh(x) = \frac{1}{2}( e^x + e^{-x} )$

    when x tends to infinity ,

    $\displaystyle \cosh(x) = \frac{e^x}{2} $

    we'll have $\displaystyle \lim_{x\to\infty} 2 \frac{e^x}{x^2} = \infty$
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  3. #3
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    im sorry but i dont really get what u mean.

    how does the coshx = e^x / 2 come about ?
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  4. #4
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    $\displaystyle coshx = \frac{e^x + e^{-x}}{2} $. Let x tend to infinity and the $\displaystyle e^{-x} \rightarrow 0 $, while the rest of the expression increases beyond all bounds. The $\displaystyle x^2 $ also increases beyond all bounds, which is why the entire expression approaches infinity.
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  5. #5
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    Quote Originally Posted by simplependulum View Post
    when x tends to 0 , $\displaystyle x^2 \cosh(\frac{1}{x}) $ tends to infinity .

    Since $\displaystyle \cosh(x) = \frac{1}{2}( e^x + e^{-x} )$

    when x tends to infinity ,

    $\displaystyle \cosh(x) = \frac{e^x}{2} $

    we'll have $\displaystyle \lim_{x\to\infty} 2 \frac{e^x}{x^2} = \infty$

    okay ive got that. but how did the $\displaystyle 2 \frac{e^x}{x^2} = \infty$ come about ?

    Thanks for the help guys
    Last edited by defcon3; Aug 19th 2009 at 10:33 PM.
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