\lim_{x\to0}\frac{\sqrt[m]{cos\alpha x}}{x^2}=\lim_{x\to0}[(\frac{1+cos\alpha x -1}{x^2})^\frac{x^2}{cos\alpha x -1}]^\frac{(cos\alpha x-1) \alpha^2}{(\alpha x)^2m}=e^{\frac{\alpha^2}{-2m}}