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
On
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 $-\cos a$.
On
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 $$\begin{aligned}L &= \lim_{x \to 0}\frac{f(a, x)}{x^{2}}\\ &= \lim_{x \to 0}\frac{2\cos(a + x)(\cos x - 1)}{x^{2}}\\ &= -2\lim_{x \to 0}\frac{\cos(a + x)(1 - \cos x)}{x^{2}}\\ &= -2\lim_{x \to 0}\frac{\cos(a + x)(1 - \cos^{2} x)}{x^{2}(1 + \cos x)}\\ &= -2\lim_{x \to 0}\cos(a + x)\cdot\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{(1 + \cos x)}\\ &= -2\cos a\cdot 1\cdot\frac{1}{2} = -\cos a\end{aligned}$$
l'Hospital and using continuity of trigonometric functions:
$$\lim_{x\to 0}\frac{\cos(a+2x)-2\cos(a+x)+\cos a}{x^2}\stackrel{\text{l'H}}=\lim_{x\to 0}\frac{-\sin(a+2x)+\sin(a+x)}{x}\stackrel{\text{l'H}}=$$
$$=\lim_{x\to 0}\left(-2\cos(a+2x)+\cos(a+x)\right)=-\cos a$$