Evaluating $\lim_{x\to0}{\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}}}$ without L'Hopital's rule I'm currently struggling with this task:
$$\lim_{x\to0}{\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}}}$$
By now I have only come up with the idea to use the formula of sum of cubes, so, the numerator would become $\cos{4x}-\cos{5x}$ and the denominator would be $3(1-\cos{3x})$:
$$\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}} = \frac{(\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}})*(\sqrt[3]{cos^2{4x}} + \sqrt[3]{\cos{4x}}*\sqrt[3]{\cos{4x}} + \sqrt[3]{\cos^2{5x}})}{(1-\cos{3x})*(\sqrt[3]{cos^2{4x}} + \sqrt[3]{\cos{4x}}*\sqrt[3]{\cos{4x}} + \sqrt[3]{\cos^2{5x}})} = \frac{\cos{4x}-\cos{5x}}{3(1-\cos{3x})}$$
In this expression it looks like $\cos{4x}-\cos{5x}$ and $(1-\cos{3x})$ increase with the same speed (if I can say it this way) and the limit is likely to be $\frac{1}{3}$.
Still I have no idea how to prove that without using L'Hopital's rule which is prohibited by the task.

Adding and subtracting $1$ in the numerator we can rewrite the expression under limit as $$\frac{1-\sqrt [3]{\cos 5x}}{1-\cos 3x}-\frac{1-\sqrt[3]{\cos 4x}} {1-\cos 3x}$$ and the first fraction above can be further rewritten as $$\frac{1-\sqrt[3]{\cos 5x}}{1-\cos 5x}\cdot\frac{1-\cos 5x}{(5x)^2}\cdot\frac{5^2}{3^2}\cdot\frac{(3x)^2}{1-\cos 3x}$$ so that it tends to $(1/3)(1/2)(25/9)(2)=25/27$. Similarly second fraction tends to $16/27$ and hence the desired limit is $9/27=1/3$.
The following standard limits are used in the above approach $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1},\,\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}$$
