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lemon14 · Jan 10, 2012, 03:09 PM

\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 ?

ebaines · Jan 10, 2012, 03:39 PM

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

$ \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. $

lemon14 · Jan 11, 2012, 03:30 AM

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...

ebaines · Jan 11, 2012, 07:55 AM

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.