Calculate the following limit
$\displaystyle \lim_{x \to 0} \ (x^2) cosh (1/x) $
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 $\displaystyle \lim_{x\to\infty} 2 \frac{e^x}{x^2} = \infty$
$\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.