Resembling resemblance?

The given integral can be computed as follows: $\int^{\infty}_0\frac{\sin(x^n)}{x}dx=\frac1n\int^{\infty}_0\frac{(nx^{n-1})\sin(x^n)}{x^n}dx$ With an appropriate substitution, the given integral becomes the well known Dirichlet Integral. So the limit $\lim\limits_{n\to\infty}\frac{\pi}{2n}=0$