I found the value of this correctly as $\frac{\pi}{2}$; $$\lim_{n \to \infty}\frac{\int_\frac{1}{n+1}^\frac{1}{n}tan^{-1}(nx) dx}{\int_\frac{1}{n+1}^\frac{1}{n}sin^{-1}(nx) dx}$$ But the solution I saw to this makes a substitution of $nx=t$, thereby changing $dx$ to $\frac{dt}{n}$ and changing the arguments of arctan and arcsine to t. All this is fine, but they also change the limits in both integrals to ranging from $\frac{n}{n+1}$ to $1$.
Why and how is this done?
As you have substituted $t = nx$, you need to change the limits corresponding to the variable $t$
At $x = \frac1n $, $t = n\frac1n = 1$
and at $x = \frac{1}{n+1}$, $t = \frac{n}{n+1}$