Let $t,x \in \mathbb{R}$. I need to show that $\lim_{t \to \pm \infty} \ |\cos(x+it)| = \infty$ and since I didn't do Calculus for quite some time I wanted to ask whether my solution is correct. (I only used some basic facts about hyperbolic functions.):
$ \lim_{t \to \pm \infty} \ |\cos(x+it)| $
$ = \lim_{t \to \pm \infty}|\cos(x)\cos(it)-\sin(x)\sin(it)|$
$ = \lim_{t \to \pm \infty}|\cos(x)\cosh(t)-\sin(x)\sinh(t)/(-i)|$
$ = \lim_{t \to \pm \infty}|\cos(x)\cosh(t)+i\sin(x)\sinh(t)|$
By definition of the absolute value in $\mathbb{C}$:
$ = \lim_{t \to \pm \infty} \cos^2(x)\cosh^2(t)-\sin^2(x)\sinh^2(t)$
By use of elementary identities:
$ = \lim_{t \to \pm \infty}[1-\sin^2(x)][1-\sinh^2(t)] - \sin^2(x)\sinh^2(t)$
$ = \lim_{t \to \pm \infty} 1-\sinh^2(t) - \sin^2(x) + \sin^2(x)\sinh^2(t) - \sin^2(x)\sinh^2(t)$
$ = \lim_{t \to \pm \infty} \ 1-\sinh^2(t) - \sin^2(x)$
$ = \lim_{t \to \pm \infty} \cosh^2(t) - \sin^2(x)$
Since $\lim_{t \to \pm \infty} \cosh(t) = \infty$:
$ = \infty$
Thanks for any answers.