# Finding the limit

• Aug 23rd 2009, 12:10 PM
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?
• Aug 23rd 2009, 12:21 PM
Chris L T521
Quote:

Originally Posted by sweetadam
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, $\displaystyle x\to0^+\implies u\to\infty$

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