Recall that
[tex]\sin(x) = \displaystyle \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}[/tex]
which converges for all real x. (One can show this with the ratio test, for instance.)
Then
[tex]\sin\left(x^4\right) = \displaystyle \sum_{n=0}^\infty (-1)^n \frac{\left(x^4\right)^{2n+1}}{(2n+1)!} = \boxed{\sum_{n=0}^\infty (-1)^n \frac{x^{8n+4}}{(2n+1)!}}[/tex]
By the ratio test,
[tex]\displaystyle \lim_{n\to\infty} \left|\frac{(-1)^{n+1} x^{8(n+1)+4}}{(2(n+1)+1)!} \times \frac{(2n+1)!}{(-1)^n x^{8n+4}}\right| \\\\ = x^8 \lim_{n\to\infty} \frac{(2n+1)!}{(2n+3)!} \\\\ = x^8 \lim_{n\to\infty} \frac1{(2n+3)(2n+2)} = 0[/tex]
