Solving Trigonometric Limits compute the following limit:
$$\lim \frac{\cos(a+2x) - 2cos(a+x) + \cos(a)}{x^2}\   \mbox{as}\ x\to0.$$
How would I go about solving this problem? I have attempted to use trig identities (addition of angles) to try to simplify the problem, but it only seemed to make it worse...
but without used ´Hospital

Note that the numerator of the expression under limit operation can be expressed as $$\begin{aligned}f(a, x) &= \cos (a + 2x) - 2\cos(a + x) + \cos a\\
&= \cos (a + 2x) + \cos a - 2\cos(a + x)\\
&= 2\cos(a + x)\cos x - 2\cos(a + x)\\
&= 2\cos(a + x)(\cos x - 1)\end{aligned}$$ and hence the desired limit $L$ can be evaluated as $$\begin{aligned}L &= \lim_{x \to 0}\frac{f(a, x)}{x^{2}}\\
&= \lim_{x \to 0}\frac{2\cos(a + x)(\cos x - 1)}{x^{2}}\\
&= -2\lim_{x \to 0}\frac{\cos(a + x)(1 - \cos x)}{x^{2}}\\
&= -2\lim_{x \to 0}\frac{\cos(a + x)(1 - \cos^{2} x)}{x^{2}(1 + \cos x)}\\
&= -2\lim_{x \to 0}\cos(a + x)\cdot\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{(1 + \cos x)}\\
&= -2\cos a\cdot 1\cdot\frac{1}{2} = -\cos a\end{aligned}$$
