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!

$\displaystyle \begin{aligned}L &= \lim_{x \to 0}\frac{\sqrt[m]{\cos x} - \sqrt[n]{\cos x}}{x^{2}}\\
&= \lim_{x \to 0}\frac{\sqrt[m]{1 - 2\sin^{2}(x/2)} - \sqrt[n]{1 - 2\sin^{2}(x/2)}}{x^{2}}\\
&= \lim_{x \to 0}\frac{\sqrt[m]{1 - 2\sin^{2}(x/2)}}{x^{2}} - \frac{\sqrt[n]{1 - 2\sin^{2}(x/2)}}{x^{2}}\\
&= \lim_{x \to 0}\frac{\sqrt[m]{1 - 2\sin^{2}(x/2)} - 1}{x^{2}} - \frac{\sqrt[n]{1 - 2\sin^{2}(x/2)} - 1}{x^{2}}\\
&= \lim_{x \to 0}\frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/m} - 1^{1/m}}{x^{2}} - \frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/n} - 1^{1/n}}{x^{2}}\\
&= \lim_{x \to 0}\frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/m} - 1^{1/m}}{-2\sin^{2}(x/2)}\cdot\frac{-2\sin^{2}(x/2)}{x^{2}}\\
&\,\,\,\,\,\,\,\,\,\,\,- \frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/n} - 1^{1/n}}{-2\sin^{2}(x/2)}\cdot\frac{-2\sin^{2}(x/2)}{x^{2}}\\
&= \lim_{x \to 0}\frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/m} - 1^{1/m}}{(1 - 2\sin^{2}(x/2)) - 1}\cdot\frac{-2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x^{2}}\\
&\,\,\,\,\,\,\,\,\,\,\,- \frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/n} - 1^{1/n}}{(1 -2\sin^{2}(x/2)) - 1}\cdot\frac{-2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x^{2}}\\
&= \frac{1}{m}\cdot 1^{(1 - m)/m}\cdot (-2)\cdot\frac{1}{4} - \frac{1}{n}\cdot 1^{(1 - n)/n}\cdot (-2)\cdot\frac{1}{4}\\
&= \frac{1}{2n} - \frac{1}{2m}\end{aligned}$
We have used standard limits $$\lim_{y \to a}\frac{y^{n} - a^{n}}{y - a} = na^{n - 1}$$ where $y = 1 - 2\sin^{2}(x/2), a = 1$ and $n$ is $1/m$ at one place and $1/n$ at another place. Also we make use of the $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$.
