Here is how I'd do with maybe more of an intuitive approach (all of this can be made more rigorous using taylor expansion for example)
$\sin(x)$ goes like $x$ in the neighborhood of the origin,
using that you have (for example):
$\csc(3/n^3)$ behaves like $n^3/3$ for very large $n$
Similarly, $\cos(x)$ can be approximated by $1$ around the origin thus $\cot(1/x)$ goes like $1/(1/x)$ with very large $x$ (ie. goes like $x$).
using that kind of approach, it is rather straightforward to obtain the solution. Hope this helps.
EDIT: hm, I did it for fun on the side and you might have forgotten something as, if I'm not mistaken, the limit does not converge. (Leading term in the numerator goes like $n^4$ and leading term in the denominator goes like $n^3$, to be more precise the leading term of the expansion at infinity is $\frac23 n$). you might want to check your equation but it doesn't change that you can still use that kind of approach for trigonometric functions.