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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?

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  • $\begingroup$ $\sinh(2z) = 2\sinh(z)\cosh(z)$. $\endgroup$ – Cameron Williams Jan 25 '16 at 3:08
  • $\begingroup$ Thanks! I ended up with the limit being equivalent to 2. $\endgroup$ – Tunk Jan 25 '16 at 3:18
  • $\begingroup$ @JaySaunders Just to show another way, I went ahead and solved it by using the exponential form. I often find this way to be superior in general, because I don't have to memorize a bunch of trig rules! Your answer matches mine though, and Wolfram Alpha confirms this :) $\endgroup$ – Brevan Ellefsen Jan 25 '16 at 3:29
  • $\begingroup$ @BrevanEllefsen Yeah I agree that exponential form is superior. I was having trouble breaking it down. However, thanks to you, I now know how to get by that road block in the future! $\endgroup$ – Tunk Jan 25 '16 at 3:57
  • $\begingroup$ @JaySaunders sure thing, glad I could help! Also, since your account seems new to Math.SE, I figured I'd let you know that to choose an answer click the "check" mark next to it... doing so is on one hand a way to "close" a question in a sense, as people see it has an accepted answer and thus the OP is satisfied. It also gives both you and the person who answered a bit of cred (useful in your case especially, since you don't get many basic privileges until you get to $100 rep) $\endgroup$ – Brevan Ellefsen Jan 25 '16 at 4:44
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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.

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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.

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  • $\begingroup$ The question has tag Complex analysis. In the complex plane, $e^{-2z}$ has no limit as $z\to\infty$. $\endgroup$ – Julián Aguirre Jan 25 '16 at 15:05
  • $\begingroup$ @JuliánAguirre So you're saying none of the methods used are correct? $\endgroup$ – Tunk Jan 25 '16 at 18:19
  • $\begingroup$ If as the question suggests $z\to\infty$ in the complex plane, then the limit does not exist. The function has poles at $(\pi/2+k\,\pi)i$, $k\in\mathbb{Z}$, and $\infty$ is an accumulation point of poles. It is not an isolated singularity. $\endgroup$ – Julián Aguirre Jan 26 '16 at 10:24
  • $\begingroup$ @JuliánAguirre. It is sure that I missed the complex-analysis tag. I thought in was in the real domain. $\endgroup$ – Claude Leibovici Jan 26 '16 at 10:50
  • $\begingroup$ @JuliánAguirre Can you break the problem down in another way, or explain your reasoning at a more basic level? Reason being, we've only just talked about analyticity and Cauchy-Riemann conditions in my class. Thanks $\endgroup$ – Tunk Jan 26 '16 at 15:22
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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.

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