Past exam of the mathematical methods II course of the University of Oxford / 2022 / 1 / ii

For $x>2$, writeSince $|\sin x|\le 1$, the denominator factor tends to one, and $\ln x/x\to 0$. Therefore the limit is $\boxed{0}$. The function is eventually negative, so it approaches zero from below. Oscillation of $\sin x$ does not obstruct this limit, because it is bounded while $x$ grows.