Evaluate $\lim_{x\to 0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2x},m\in \mathbb{N}$ 
Evaluate $$\lim_{x\to 0}\frac{\sqrt[m]{\cos\alpha x}-\sqrt[m]{\cos\beta x}}{\sin^2x},m\in \mathbb{N}$$

I used L'Hospital's rule, but that didn't work. Could Taylor series be used? I don't know how to use it with irrational functions.

This can be done via the use of following standard limits $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1},\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\lim_{x \to 0}\frac{\sin x}{x} = 1$$ Let us proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt[m]{\cos \alpha x} - \sqrt[m]{\cos \beta x}}{\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt[m]{\cos \alpha x} - \sqrt[m]{\cos \beta x}}{x^{2}}\cdot\frac{x^{2}}{\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt[m]{\cos \alpha x} - \sqrt[m]{\cos \beta x}}{x^{2}}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{(1 - \sqrt[m]{\cos \beta x}) - (1 - \sqrt[m]{\cos \alpha x})}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{(1 - \sqrt[m]{\cos \beta x})}{x^{2}} - \frac{(1 - \sqrt[m]{\cos \alpha x})}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{1 - \sqrt[m]{\cos \beta x}}{1 - \cos \beta x}\cdot\frac{1 - \cos \beta x}{x^{2}} - \frac{1 - \sqrt[m]{\cos \alpha x}}{1 - \cos \alpha x}\cdot\frac{1 - \cos \alpha x}{x^{2}}\notag\\
&= \lim_{t \to 1}\frac{1 - t^{1/m}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos \beta x}{x^{2}} - \lim_{t \to 1}\frac{1 - t^{1/m}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos \alpha x}{x^{2}}\notag\\
&= \frac{1}{m}\left(\lim_{x \to 0}\frac{1 - \cos \beta x}{x^{2}} - \lim_{x \to 0}\frac{1 - \cos \alpha x}{x^{2}}\right)\notag\\
&= \frac{1}{m}\left(\lim_{x \to 0}\beta^{2}\cdot\frac{1 - \cos (\beta x)}{(\beta x)^{2}} - \lim_{x \to 0}\alpha^{2}\cdot\frac{1 - \cos (\alpha x)}{(\alpha x)^{2}}\right)\notag\\
&= \frac{1}{m}\left(\frac{\beta^{2}}{2} - \frac{\alpha^{2}}{2}\right)\notag\\
&= \frac{\beta^{2} - \alpha^{2}}{2m}\notag
\end{align}
