<s>[INST] 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 "$-4$"
 [/INST] 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*}
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