Does the limit : $\lim _{x\to \infty }\frac{\ln x^{\frac{1}{3}}}{\sin x}$ exist

Let $f \left(x\right) = \frac{\ln {x}^{1/3}}{\sin \left(x\right)}$. We have

\begin{equation}\liminf\limits _{\substack{x \rightarrow \infty \\ x \notin {\pi} \mathbb{Z} }} f \left(x\right) \leqslant \lim _{n \rightarrow \infty } f \left(n {\pi}-\arcsin \left(1/n\right)\right) = \lim _{n \rightarrow \infty } \frac{\ln {\left(n {\pi}-\arcsin \left(1/n\right)\right)}^{1/3}}{{-1}/n} =-\infty \end{equation}

and

\begin{equation}\limsup\limits _{\substack{x \rightarrow \infty \\ x \notin {\pi} \mathbb{Z} }} f \left(x\right) \geqslant \lim _{n \rightarrow \infty } f \left(n {\pi}+\arcsin \left(1/n\right)\right) = \lim _{n \rightarrow \infty } \frac{\ln {\left(n {\pi}+\arcsin \left(1/n\right)\right)}^{1/3}}{1/n} =+\infty \end{equation}

Hence $f \left(x\right)$ has no limit when $x \rightarrow \infty , x \notin {\pi} \mathbb{Z}$.