# Thread: Limit Problem

1. ## Limit Problem

Calculate the following limit

$\lim_{x \to 0} \ (x^2) cosh (1/x)$

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

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

when x tends to infinity ,

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

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

3. im sorry but i dont really get what u mean.

how does the coshx = e^x / 2 come about ?

4. $coshx = \frac{e^x + e^{-x}}{2}$. Let x tend to infinity and the $e^{-x} \rightarrow 0$, while the rest of the expression increases beyond all bounds. The $x^2$ also increases beyond all bounds, which is why the entire expression approaches infinity.

5. Originally Posted by simplependulum
when x tends to 0 , $x^2 \cosh(\frac{1}{x})$ tends to infinity .

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

when x tends to infinity ,

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

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

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

Thanks for the help guys