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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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 $\displaystyle u=\frac{1}{x}$. Thus, $\displaystyle x\to0^+\implies u\to\infty$Quote:
Originally Posted by sweetadam
Therefore, $\displaystyle \lim_{x\to0^+}x^2\cosh\left(\tfrac{1}{x}\right)=\l im_{u\to\infty}\frac{\cosh\left(u\right)}{u^2}=\tf rac{1}{2}\lim_{u\to\infty}\frac{\sinh u}{u}=\tfrac{1}{2}\lim_{u\to\infty}\cosh u=\infty$
Let $\displaystyle u=\frac{1}{x}$. This time, $\displaystyle x\to0^-\implies u\to-\infty$
Therefore , $\displaystyle \lim_{x\to0^-}x^2\cosh\left(\tfrac{1}{x}\right)=\lim_{u\to-\infty}\frac{\cosh\left(u\right)}{u^2}=\tfrac{1}{2 }\lim_{u\to-\infty}\frac{\sinh u}{u}=\tfrac{1}{2}\lim_{u\to-\infty}\cosh u=\infty$
Therefore, $\displaystyle \lim_{x\to 0}x^2\cosh\left(\tfrac{1}{x}\right)=\infty$
