Evaluating a Rather Complicated Limit with a Simple Solution: $\lim_{x \to 0} {\cos(3x)-\cos(x) \over x^2}$ 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 "$-4$"
 A: 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.
A: $${\displaystyle \cos(3\theta )=4\cos ^{3}x -3\cos x}$$
$$ {4\cos ^{3}x -3\cos x -\cos x \over x^2}=\frac{4\cos x(\cos^2 x-1)}{x^2}=-4\cos x(\frac{\sin^2x}{x^2})$$
so the limit is $-4$
A: 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*}
A: Use L'Hospital's Rule directly,
$$\lim_{x \to 0}\frac{\cos 3x −\cos x}{x^2}
=\lim_{x \to 0}\frac{-3\sin 3x +\sin x}{2x}
=\lim_{x \to 0}\frac{-9 \cos 3x+\cos x}{2}=\lim_{x \to 0}\frac{-9+1}{2}=-4$$
