Can't understand how to solve limit like this:
$$\lim_{x \to 0} \frac{\cos3x-\cos x}{\tan2x^2}$$
My attempt is:
$$\lim_{x \to 0} \frac{\cos3x-\cos x}{\tan2x^2}=\lim_{x \to 0} \frac{\cos3x}{\tan2x^2}- \frac{\cos x}{\tan2x^2}=\lim_{x \to 0}\frac{1}{\tan2x^2}\left(\cos3x-\cos x \right) = \lim_{x \to 0} \dots$$
And could you explain me how to do next step?
(I know I have to get rid of trigonometry but I don't know how to do this.)
[Math] How to find $\lim_{x \to 0} \frac{\cos3x-\cos x}{\tan 2x^2}$
calculuslimitsreal-analysistrigonometry
Best Answer
We have $\cos(3x) - \cos(x) = -2\sin(2x)\sin(x)$. Hence, $$\dfrac{\cos(3x) - \cos(x)}{\tan(2x^2)} = -2 \dfrac{\sin(2x) \sin(x)}{\tan(2x^2)} = -2 \dfrac{\sin(2x)}{2x} \cdot \dfrac{\sin(x)}{x} \cdot \dfrac{2x^2}{\tan(2x^2)}$$ Now let $x \to 0$ and recall the following limits: $\lim_{y \to 0}\dfrac{\sin(y)}y = 1$ and $\lim_{y \to 0}\dfrac{\tan(y)}y = 1$.