I've already been able to show that $\frac{\sin(x^n)}{x^n}\rightarrow 0$ pointwise if $x>1$, and (using L'Hospital's rule) that $\frac{\sin(x^n)}{x^n}\rightarrow 1$ pointwise on $(0,1)$. But in order to show that $\lim_{n\to\infty}\int_0^\infty\frac{\sin(x^n)}{x^n}\,dx = 1$ I'd need to show that $\lim_{n\to\infty}\int_0^\infty\frac{\sin(x^n)}{x^n}\,dx = \int_0^\infty\lim_{n\to\infty}\frac{\sin(x^n)}{x^n}\,dx$.
I'm having trouble finding the right bounding to apply the dominated convergence theorem.
Any thoughts would be greatly appreciated.
Thanks.