Limit Problem

**defcon3** · August 19th 2009, 01:44 AM

Calculate the following limit

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

**simplependulum** · August 19th 2009, 02:37 AM

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 $\lim_{x\to\infty} 2 \frac{e^x}{x^2} = \infty$

**defcon3** · August 19th 2009, 02:58 AM

im sorry but i dont really get what u mean.

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

**JG89** · August 19th 2009, 06:00 AM

$coshx = \frac{e^x + e^{-x}}{2}$ . 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.

**defcon3** · August 19th 2009, 10:18 PM

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

Thread: Limit Problem

LinkBack

Thread Tools

Display

Limit Problem

Similar Math Help Forum Discussions

I have a problem in understanding a problem about the limit of a function.

limit question on a limit comparison test problem.

Another Limit Problem

Limit, Limit Superior, and Limit Inferior of a function

Limit Problem

Search Tags