For large n:

[e^(-n) + sin(n) - n^(1/3)]/[2cos(n) - n^(5/6)] -> 1/n^(1/2).

Why?

e^(-n) -> 0 and sin(n) and cos(n) become indeterminate, but vary between -1 and 1. However both n^(1/3) and n^(5/6) get very large for large n, so these will be the dominant terms. Thus

[e^(-n) + sin(n) - n^(1/3)]/[2cos(n) - n^(5/6)] -> -n^(1/3) / -n^(5/6) = 1/n^(1/2)

-Dan