$n,m\in \mathbb N$
$\sinh^{-1}\underbrace{\Big(\frac{(-1)^{m-4n}(1-n^2)}{3mn^2-5n^2+9nm-15n}\Big)}_A=\sinh^{-1}\Big(\frac{(-1)^m(1-n)(1+n)}{n(n+3)(3m-5)}\Big)$
$$\lim_{n,m\to\infty}\sin^{-1}(A)=?$$
$\;\sinh\;$ and $\;\sinh^{-1}$ are bijective $\implies \min\{\sinh^{-1}\} = \sin^{-1}(infA)$
$\Big|\frac{(-1)^m(1-n)(1+n)}{n(n+3)(3m-5)}\Big|\stackrel{m>1, n\rightarrow\infty}{=} \frac{n+1}{n(n+3)(3m-5)}\small|(-1)^m(1-n)\small|$
At this point, I believe I need to either apply L'Hospital's rule or make something from this: $\ln\Big(A^2+\sqrt{A^2+1}\Big)$. What should I do next?
The factorization of $A$ allows you to compute the double limit because $A$ can be written as the product of two sequences, the first in terms of $n$ and the second in terms of $m$. $$\lim_{n\to \infty}\frac{(1-n)(1+n)}{n(n+3)}=-1 $$ and $$\lim_{m\to \infty}\frac{(-1)^m}{3m-5}=0 $$ Therefore, $$\lim_{n,m\to\infty}A=-1\cdot0=0 $$ and using the continutity of $\sinh^{-1}(x)$ $$\lim_{n,m\to\infty}\sinh^{-1}A=\sinh^{-1}(0)=0 $$