Evaluating a Rather Complicated Limit with a Simple Solution: $\lim_{x \to 0} {\cos(3x)-\cos(x) \over x^2}$

Question

SE,

I've encountered an interesting question while practicing evaluation of limits. The following is the expression to be evaluated:

$$\lim_{x \to 0} {\cos(3x)-\cos(x) \over x^2}$$

I've tried applying the double angle formula to $\cos(3x)$ by taking it as $\cos(2x+x)$ but the expression just gets messier as I go forward. I would thus greatly appreciate a push in the right direction!

The given solution is "$-4$"

Answer 1

In case it's required, it's not too hard to show Abdallah's identity: \begin{align*} \lim_{x\to 0} \frac{1-\cos x}{x^2} &= \lim_{x\to 0} \frac{1-\cos x}{x^2}\cdot \frac{1+\cos x}{1+\cos x} \\ &= \lim_{x\to 0} \frac{1-\cos^2 x}{x^2}\cdot \frac{1}{1+\cos x} \\ &= \lim_{x\to 0} \frac{\sin^2 x}{x^2}\cdot \frac{1}{1+\cos x} \\ &= \left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2 \cdot\lim_{x\to 0} \frac{1}{1+\cos x} \\ &= 1^2 \cdot \frac{1}{1+1} = \frac{1}{2} \end{align*}

Now you can use the old add-and-subtract-the-same-thing trick: \begin{align*} \lim_{x\to 0}\frac{\cos 3x - \cos x}{x^2} &=\lim_{x\to 0}\frac{\cos 3x -1+1- \cos x}{x^2} \\ &= \lim_{x\to 0} \frac{1-\cos x}{x^2} - \lim_{x\to 0} \frac{1-\cos 3x}{x^2} \\ &= \lim_{x\to 0} \frac{1-\cos x}{x^2} - 9\lim_{3x\to 0} \frac{1-\cos 3x}{(3x)^2} \\ &= \frac{1}{2} -9 \cdot \frac{1}{2} = -4 \end{align*}

Answer 2

Using the follow formula: $$\cos 3x-\cos x=-2\sin 2x \sin x$$ \begin{align} \lim_{x\to 0}\frac{\cos 3x-\cos x}{x^2}=\lim_{x\to 0}\frac{-2\sin 2x \sin x}{x^2}=-4 \end{align} Or,you can use L'Hospital directly.

Answer 3

$${\displaystyle \cos(3\theta )=4\cos ^{3}x -3\cos x}$$ $$ {4\cos ^{3}x -3\cos x -\cos x \over x^2}=\frac{4\cos x(\cos^2 x-1)}{x^2}=-4\cos x(\frac{\sin^2x}{x^2})$$ so the limit is $-4$

Answer 4

Use L'Hospital's Rule directly, $$\lim_{x \to 0}\frac{\cos 3x −\cos x}{x^2} =\lim_{x \to 0}\frac{-3\sin 3x +\sin x}{2x} =\lim_{x \to 0}\frac{-9 \cos 3x+\cos x}{2}=\lim_{x \to 0}\frac{-9+1}{2}=-4$$