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

Consider first $$\cos (y+x)-\cos y=\cos y\cdot(\cos x-1)-\sin x\sin y=-2\cos y\cdot\sin^2\frac x2-\sin x \sin y$$
And $$\sin(a+x)-\sin a=\sin a\cdot(\cos x-1)+\cos a\sin x=-2\sin a\cdot\sin^2 \frac x2+\cos a\sin x$$
Then (expressing the numerator as a double-difference ie a difference of differences) $$\cos (a+2x)-2\cos (a+x)+\cos a = (\cos (a+2x)-\cos(a+x))-(\cos (a+x)-\cos a)$$
We apply the first identity to the two components of this to obtain $$-2\cos(a+x)\cdot \sin^2 \frac x2-\sin x \cdot\sin(a+x)+2\cos a\sin^2\frac x2+\sin x\sin a =$$using the second identity too $$=2\sin^2\frac x2\left(2\cos a\sin^2\frac x2+\sin x\sin a \right)-\sin x\left(-2\sin a\sin^2 \frac x2+\cos a\sin x\right)$$
Now the basic limit here is $$\lim _{x\to 0} \frac {\sin x}{x}=1$$ and on dividing by $x^2$ and taking the limit, the terms in $\sin^4 \frac x2$ and $\sin x\sin^2\frac x2$ go to zero, leaving the final term which goes to $-\cos a$.
