What is the limit of $cosh(2x)csch(5x)$ as $x \to +\infty$

Question

The problem asks for the $\lim f(x)$ as $x \to +\infty$ where $$f(x) = \cosh(2x)\operatorname {csch}(5x)$$ According to the limit calculators in the internet (e.g. Wolfram Alpha), the limit should be equal to $0$. This was also backed up when I checked the graph of $f(x)$.

I've tried converting $f(x)$ into its "exponential form" or whatever its called. In doing so, I got 3 possible alternative values. The first one I got is, $$\dfrac{ e^{2x} + e^{-2x} }{ e^{5x} - e^{-5x} }$$ by using the values of $\cosh(2x) = \dfrac{ e^{2x} + e^{-2x} }{2}$ and $\operatorname{csch}(5x) = \dfrac{2}{e^{5x} - e^{-5x}}$

I tried substituting $+\infty$ directly but got an $\dfrac{\infty}{\infty}$ which is indeterminate. So I tried simplifying it even more by adding the numerator and the denominator and getting the following as my second value:$$e^{3x}\dfrac{e^{4x}+1}{e^{5x}-1}$$ Then, I simplified it even more by dividing $\dfrac{e^{5x}}{e^{5x}}$ and got this as the final value: $$e^{2x}\dfrac{1+e^{-4x}}{1-e^{-5x}}$$ But by substituting the final value that I got, I always get $+\infty$ as its limit. I also checked limits for the values I got and the only one that has the same limit as the initial function, is the first one. The limits of the other two are both $+\infty$. So, I clearly got something wrong after the first one.

So. my questions are:

1. How do I solve for the $\lim f(x)$ as $x\to+\infty$?
2. If what I was doing was in the right direction, what did I do wrong and how to finish it?

Answer 1

Using equivalents, when $x$ is large $$\cosh(2x) = \dfrac{ e^{2x} + e^{-2x} }{2}\sim \dfrac{ e^{2x}} 2$$ $$\operatorname{csch}(5x) = \dfrac{2}{e^{5x} - e^{-5x}}\sim \dfrac2{ e^{5x}} $$ and the product is then $\sim e^{-3x} \to 0$

Answer 2

You have made a mistake in simplifying $\frac{e^{2x}+e^{-2x}}{e^{5x}-e^{-5x}}$: $$\frac{e^{2x}+e^{-2x}}{e^{5x}-e^{-5x}} = \frac{e^{2x}+\frac{1}{e^{2x}}}{e^{5x}-\frac{1}{e^{5x}}}=\frac{e^{4x}+1}{e^{10x}-1}\cdot\frac{e^{5x}}{e^{2x}}$$ Now as $x\to\infty$ $$\frac{e^{4x}+1}{e^{10x}-1}\sim\frac{e^{4x}}{e^{10x}}=e^{-6x}$$ and so $e^{-6x}\cdot e^{3x} = e^{-3x}\to 0.$