<s>[INST] 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.
 [/INST] You can use the Taylor Expansion of $\cos(x)$.
The numerator is $\dfrac{(5x)^2-(4x)^2}{2!} + O\left({x^4}\right)$
The denominator is $\dfrac{(3x)^2}{3\cdot2!} + O\left({x^4}\right)$
Dividing by $x^2$ and using $3^2+4^2=5^2$ gives the limit as $1/3$.
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