# Math Help - Limit problem for hyperbolic functions

1. ## Limit problem for hyperbolic functions

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?

2. 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?

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

$\chi$ $\sigma$

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