Evaluating a complex limit

Question

I would love some advice on how to approach the following limit:

$$\lim_{z\to \infty} \frac{\sinh(2z)}{\cosh^2(z)}$$

or let $z= \dfrac{1}{t}$

then $$\lim_{t\to 0} \frac{\sinh(\dfrac{2}{t})}{\cosh^2(\dfrac{1}{t})}$$

I've approached it using Taylor series and exponential forms, but to no avail. Any advice regarding how to proceed?

Answer 1

We know that $\cosh(z)=\cos(i\,z)$, so that $\cosh z=0$ when $i\,z=\pi/2+k\,\pi$, $k\in\mathbb{Z}$. The denominator in the definition of $f$ vanishes at a sequence of points converging to (complex) $\infty$. This already is a difficulty in defining $\lim_{z\to\infty}f(z)$. Now consider what happens if we approach $\infty$ through different roads. If $z=x$ is real and $x\to+\infty$, then the limit is $2$ as shown in the previous answers. But if $x$ is real and $x\to-\infty$, then the limit is $-2$. This shows that $\lim_{z\to\infty}f(z)$ does not exist, when the limit is taken in he complex plane.

Answer 2

Using Cameron Williams's comment probably makes the solution faster $$\frac{\sinh(2z)}{\cosh^2(z)}=\frac{2\sinh(z)\cosh(z)}{\cosh^2(z)}=2\frac{\sinh(z)}{\cosh(z)}=2\frac{e^z-e^{-z}}{e^z+e^{-z}}=2\frac{1-e^{-2z}}{1+e^{-2z}}$$ Since $z\to \infty$, $e^{-2z}\to 0$ and then the limit.

Answer 3

If $f$ is a meromorphic function on $\mathbb{C}$ which is not rational, then $f$ has an essential singularity at $z=\infty$ (or $z = \infty$ is a limit point of poles). Either way, $$ \lim_{z\to \infty} f(z) $$ does not exist.