0 \leq \lvert \dfrac {{(cosxsinx)^5\over x^{10}}}{({sinx\over x}-{cos^2x\over x^2})^2+{cos^8x\over x^8}} \rvert = \lvert \dfrac {{(cosxsinx)^5\over x^{10}}}{ \frac{x^4(x sinx - cos^2x)^2 + cos^8x}{x^8}} \rvert \leq \frac{\frac{1}{x^{10}}}{\frac{1}{16 x^8}} = \frac{16}{x^2}