Limit problem for hyperbolic functions

**Malaclypse** · July 24th 2010, 11:23 AM

Hi all,

I wanted to know if I could have your insight on the following problem:

$\mathop {\lim }\limits_{x \to \infty } \frac{{\cosh x + \sqrt x }}{{{e^x} + \ln x}}$

I've tried to tackle it with L'Hopital, however that seems to create an endless loop. I played around with using logarithmic differentiation thinking that might open something up, however that seemed to create more of a mess than anything else.

I'm thinking perhaps there is an identity or something that I'm missing that would allow me to cancel some things out along the way. With that hyperbolic cos in the function I just keep jumping between sinh and cosh as I differentiate over and over again.

It's something simple that I'm missing I'm sure, but simple or no I'm missing it!

Thoughts?

Thanks again in advance for your help.

**chisigma** · July 24th 2010, 11:46 AM

Originally Posted by Malaclypse

Hi all,

I wanted to know if I could have your insight on the following problem:

$\mathop {\lim }\limits_{x \to \infty } \frac{{\cosh x + \sqrt x }}{{{e^x} + \ln x}}$

I've tried to tackle it with L'Hopital, however that seems to create an endless loop. I played around with using logarithmic differentiation thinking that might open something up, however that seemed to create more of a mess than anything else.

I'm thinking perhaps there is an identity or something that I'm missing that would allow me to cancel some things out along the way. With that hyperbolic cos in the function I just keep jumping between sinh and cosh as I differentiate over and over again.

It's something simple that I'm missing I'm sure, but simple or no I'm missing it!

Thoughts?

Thanks again in advance for your help.

Because is $\cosh x = \frac{e^{x} + e^{-x}}{2}$ , deviding numerator and denominator by $e^{x}$ You obtain...

$\displaystyle \lim_{x \rightarrow \infty} \frac{\cosh x + \sqrt {x}}{e^{x} + \ln x} = \lim_{x \rightarrow \infty} \frac{\frac{1}{2}\ (1+e^{-2x}) + \sqrt{x}\ e^{-x}}{1 + \ln x\ e^{-x}} = \frac{1}{2}$

Kind regards

$\chi$ $\sigma$

**Malaclypse** · July 24th 2010, 11:53 AM

Of course...I knew it was something simple. Thanks!

Thread: Limit problem for hyperbolic functions

LinkBack

Thread Tools

Display

Limit problem for hyperbolic functions

Similar Math Help Forum Discussions

Hyperbolic Limit Question

Hyperbolic functions

Is it possible to solve this "hyperbolic functions" problem?

Hyperbolic Functions

Hyperbolic Limit