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Math Help - Finding the limit

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    Finding the limit

    How do I find the limit of (x^2)(cosh(1/x) when x is approching zero, I know its is ∞ but how do I show the steps?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by sweetadam View Post
    How do I find the limit of (x^2)(cosh(1/x) when x is approching zero, I know its is ∞ but how do I show the steps?
    Let u=\frac{1}{x}. Thus, x\to0^+\implies u\to\infty

    Therefore, \lim_{x\to0^+}x^2\cosh\left(\tfrac{1}{x}\right)=\l  im_{u\to\infty}\frac{\cosh\left(u\right)}{u^2}=\tf  rac{1}{2}\lim_{u\to\infty}\frac{\sinh u}{u}=\tfrac{1}{2}\lim_{u\to\infty}\cosh u=\infty

    Let u=\frac{1}{x}. This time, x\to0^-\implies u\to-\infty

    Therefore , \lim_{x\to0^-}x^2\cosh\left(\tfrac{1}{x}\right)=\lim_{u\to-\infty}\frac{\cosh\left(u\right)}{u^2}=\tfrac{1}{2  }\lim_{u\to-\infty}\frac{\sinh u}{u}=\tfrac{1}{2}\lim_{u\to-\infty}\cosh u=\infty

    Therefore, \lim_{x\to 0}x^2\cosh\left(\tfrac{1}{x}\right)=\infty
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