Limit with Integration - 6

$|\sin(x^n)|\le1\therefore \frac{|\sin(x^n)|}{x^n}\le\frac{1}{x^n}$ $\int_1^\infty\frac{\sin(x^n)}{x^n}dx\le\int_1^\infty x^{-n}dx=\left[\frac{x^{1-n}}{n-1}\right]_\infty^1=\frac{1}{n-1}\{n>1\}$ which converges to zero, now let us focus on the following portion of domain $I=\int_0^1\frac{\sin(x^n)}{x^n}dx$ we know that: $\lim_{x\to 0}\frac{\sin(x^n)}{x^n}\to1$ and that is the region $x\in[0,1],\,\,\frac{\sin(x^n)}{x^n}\le1$ furthermore we can show that for $n\to\infty,\frac{\sin(x^n)}{x^n}\to1$ and so taking the limit first we get: $\lim_{n\to\infty}\int_0^1\frac{\sin(x^n)}{x^n}dx\to\int_0^1dx=1$ combining this with the first integral calculated and taking the limit we get our desired result.