How to evaluate limit without L'Hospital's rule, but using $\lim_{x\to 0} \dfrac{\sin x}{x} = 1$

Question

I am having hard time evaluating this limit:

$$\lim_{x\to0} \frac{\cos3x-1}{\cos2x-1}$$

I know, that this fact can help $$\lim_{x\to0} \frac{\sin(x)}{x} = 1$$

But how? We can transform the first expression to $$\lim_{x\to0} \frac{\cos3x-1}{-2\sin^2x}$$ but what should I do next? I cannot use L'Hopital's rule for solving this. Can you help?

Answer 1

$$\lim_{x \to 0} \frac{\cos 3x -1}{\cos 2x -1} = \lim_{x\to 0} \frac{2\sin^2\left(\frac {3x}{2}\right)}{2\sin^2(x)}$$ $$=\lim_{x\to 0} \frac{\frac{\sin^2\left(\frac{3x}{2}\right)}{\left(\frac{3x}{2}\right)^2}\left(\frac{3x}{2}\right)^2}{\frac{\sin^2(x)}{x^2}x^2}$$

$$=\lim_{x\to 0} \frac{\frac{9x^2}{4}}{x^2}$$ $$=\frac 94$$

Answer 2

Hint:

It is standard (from the limit of $\sin x/x$) that $$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac 12,\quad\text{so}\quad 1-\cos x\sim_0\frac{x^2}2,$$ and you may proceed by substitution.

Answer 3

$$\lim_{x\to0} \frac{\cos3x-1}{\cos2x-1}=\frac{9}{4}\lim_{x\to0} \frac{-\frac{1-\cos3x}{9x^2}}{-\frac{1-\cos2x}{4x^2}}=\frac{9}{4}$$

Answer 4

Another option is to use $1-\cos x=\frac{\sin^2x}{1+\cos x}\stackrel{x\to0}{\sim}\frac12\sin^2x\sim\frac12x^2$, so the limit is$$\lim_{x\to0}\frac{1-\cos 3x}{1-\cos 2x}=\lim_{x\to0}\frac{\frac12(3x)^2}{\frac12(2x)^2}=\frac94.$$