# How to show that the limit of cosh(z) as z goes to infinity does not exist?

I am trying to understand what does it mean to calculate a limit of complex function as z approaches infinity. It is not intuitive to me. So far when calculating a finite limit (complex) it was clear that real part approaches real part and imaginary to imaginary. In an example exercise I was asked to calculate the limit of the function exp(z)/(exp(-z)+exp(z)) and the answer simplifying the function expression to 1/2cosh(z) concluding the limit of cosh(z) as z goes to infinity does not exist, and thus niether does the limit of the function, however without further explanation. I'm not quite sure how to practically show this conclusion. I would appreciate a detailed explanation/calculation. Thanks

• The Question's wording and its tag for complex analysis suggests you might be looking for a limit at infinity in the sense of the complex plane. The only entire functions that have limits at infinity are polynomials, and among these, only constants have finite limits. – hardmath Jul 29 '16 at 20:35
• Thanks. What if i write coshz as taylor polynomial ? – breeze Jul 29 '16 at 21:09
• While $\cosh (z)$ is easily expressed as a power series related to that for $\cos (z)$ as the Accepted Answer mentions, this does not give a polynomial expression. – hardmath Jul 29 '16 at 21:17
• Thank you. May I ask, Is there a theorem stating the above conclusion that only entire functions have limits at infinity are polynomials? Please share – breeze Jul 29 '16 at 21:38
• en.wikipedia.org/wiki/Liouville%27s_theorem_(complex_analysis) – Hurkyl Aug 15 '16 at 18:24

Hint: Note that $\cos x=\cosh(ix)$. Now what happens to $\cos x$ as $x\to\infty$ along the $x$-axis?