Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$

By standard limits

  • $\frac{\sin x}x \to 1$

  • $\frac{1-\cos x}{x^2} \to \frac12$

we have that

$$\frac{\cos x-\cos(3x)}{\sin(3x^2)-\sin(x^2)}=\frac{\frac{\cos x-1+1- \cos(3x)}{x^2}} {\frac{\sin(3x^2)-\sin(x^2)}{x^2}}=\frac{-\frac{1-\cos x}{x^2}+9\frac{1- \cos(3x)}{(3x)^2}} {3\frac{\sin(3x^2)}{3x^2}-\frac{\sin(x^2)}{x^2}}\to\frac{-\frac12+\frac92}{3-1}=2$$


Hint: Use $$\sin 3a=3\sin a-4\sin^3a$$ $$\cos 3a=4\cos^3a-3\cos a$$ Edit: After substutution it is $$\lim_{x\to0}\frac{2\cos x\sin^2x}{\sin x^2\cos2x^2}=2$$


Hint: Use the factorisation formula $$\cos a -\cos b =-2\sin{a+b\over 2}\sin{a-b\over 2}$$ and $$\sin a -\sin b =2\sin{a-b\over 2}\cos{a+b\over 2}$$

$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} = \lim_{x\to0}\frac{2\color{red}{\sin 2x}\cdot \color{green}{\sin x}\cdot\color{blue}{x^2}}{\color{blue}{\sin (x^2)}\cos (2x^2)\cdot\color{red}{2x}\cdot \color{green}{x}} =2$$

Tags:

Trigonometry

Limits

Calculus

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