# Evaluating a complex limit

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?

• $\sinh(2z) = 2\sinh(z)\cosh(z)$. – Cameron Williams Jan 25 '16 at 3:08
• Thanks! I ended up with the limit being equivalent to 2. – Tunk Jan 25 '16 at 3:18
• @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 :) – Brevan Ellefsen Jan 25 '16 at 3:29
• @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! – Tunk Jan 25 '16 at 3:57