\lim_{x \to 0} \frac {\cos 5x - \cos x}{\sin 4x} = \lim_{x \to 0} \frac{-2 \sin 3x \sin 2x}{\sin 4x} = \lim_{x \to 0} = \frac {-2 \frac {\sin 3x}{3x} \times 3x \times \frac {\sin 2x}{2x} \times 2x }{ \frac{\sin 4x}{4x} \times 4x }

Why doesn't it reach any answer or what is wrong ?

First off - you need to enclose your LaTex notation between and [/MATH ] tags:

[MATH]
\lim_{x \to 0} \frac {\cos 5x - \cos x}{\sin 4x} = \lim_{x \to 0} \frac{-2 \sin 3x \sin 2x}{\sin 4x} = \lim_{x \to 0} = \frac {-2 \frac {\sin 3x}{3x} \times 3x \times \frac {\sin 2x}{2x} \times 2x }{ \frac{\sin 4x}{4x} \times 4x }


If you take the limits of the derivatives of numerator and denominator you get:


\frac {\lim _{x \to 0} (-5\sin 5x+\sin x)}{\lim_{x \to 0} (4 \cos 4x)} = \frac { 0 + 0} {4} = 0.

Well, I don't quite understand your explanation for I haven't learned about derivatives yet, that's why I tried to use the formula \lim_{x \to 0} \frac{sin x}{x} = 1 , but at the numerator there is x^2 and at the denominator only x . As long as I have one x more I suppose what I solved is not entirely correct...

From this step:


\lim_{x \to 0} (\frac {-2 \frac {\sin 3x}{3x} \times 3x \times \frac {\sin 2x}{2x} \times 2x }{ \frac{\sin 4x}{4x} \times 4x })


You can divide numerator and denominator by x:


\lim_{x \to 0} ( \frac {-2 \frac {\sin 3x}{3x} \times \frac {\sin 2x}{2x} \times 6x }{ \frac{\sin 4x}{4x} \times 4 }) = \frac {-2 \times 1 \times 1 \times 0}{1 \times 4 } = 0


It's legal to divide through by x because x is NOT equal to 0; it approaches 0 but it never actually equals 0.