Compute $\lim_{x\to 0}\dfrac{\sqrt[m]{\cos x}-\sqrt[n]{\cos x}}{x^2}$ For two positive integers $m$ and $n$, compute
$$\lim_{x\to 0}\dfrac{\sqrt[m]{\cos x}-\sqrt[n]{\cos x}}{x^2}$$  
Without loss of generality I consider $m>n$ and multiply the numerator with its conjugate. But what next? Cannot proceed further! Help please!

$$\lim_{x\to 0}\dfrac{\sqrt[m]{\cos x}-\sqrt[n]{\cos x}}{x^2}=\lim_{x\to 0}\dfrac{\sqrt[m]{1+\cos x-1}-\sqrt[n]{1+\cos x-1}}{x^2}=$$$$=\lim_{x\to 0}\dfrac{(1+\cos x-1)^{\frac{1}{m}}-1}{\cos x-1}\frac{\cos x-1}{x^2}-\lim_{x\to 0}\dfrac{(1+\cos x-1)^{\frac{1}{n}}-1}{\cos x-1}\frac{\cos x-1}{x^2}= $$$$=-\frac{1}{2m}-(-\frac{1}{2n})= \frac{1}{2n}-\frac{1}{2m}$$ 
I applied: $$\lim_{x\to 0}\frac{(1+x)^r-1}{x}=r$$ and $$\lim_{x\to 0}\frac{1-\cos x}{x^2} =\frac{1}{2}.$$
