Using $ x - \frac{x^3}{6} \lt \sin x \lt x$
for positive $x$ close to $0$ and the fact that $\displaystyle \sum_{k=1}^{n} \frac{1}{k^3}$ is convergent, we see that
your limit is the same as the limit of $\displaystyle s_n = \sum_{k=1}^{n} \frac{1}{n+k}$
One method to find this limit is to use the Harmonic series estimate as in Alex's answer.
Another method is to interpret it as a Riemann sum of the integral $\int_{0}^{1} \frac{1}{1+x} \text{ dx}$ as in Ragib's comment to Alex's answer.
I will mention a third method, which uses the Mean Value theorem. (though all three are quite similar).
Since $\frac{1}{x}$ is decreasing for positive $x$, we see using that mean value theorem that
$\frac{1}{x+1} \lt \log (1+x) - \log x \lt \frac{1}{x}$
Setting $x = n+1, n+2, \dots, 2n$ and adding, and then setting $x=2n-1, 2n-2, \dots, n$ and adding we get
$\log \left(\frac{2n+1}{n}\right) \lt s_n \lt \log \left(\frac{2n-1}{n-1}\right)$
and thus
$\lim_{n \to \infty} s_n = \log 2$