# 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. Jul 29 '16 at 20:35
• Thanks. What if i write coshz as taylor polynomial ? 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. 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 Jul 29 '16 at 21:38
• en.wikipedia.org/wiki/Liouville%27s_theorem_(complex_analysis)
– user14972
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?