$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} $ I have to find for which values of $a \in \Bbb N, a \ne 0$ the following limit exists and it is finite:
$$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2}  $$
Applying L'Hôpital's rule:
$$\frac{1-\cos (ax)}{x^2}\sim \frac{\sin (ax) \cdot a}{2x}\sim \frac{a^2}{2}$$
Then $$ \frac{\cos (\pi \cdot \frac{1-\cos (ax)}{x^2})}{x^2}  \sim   \frac{\cos (\pi \cdot \frac{a^2}{2})}{x^2}.$$
$$\cos (\pi \cdot \frac{a^2}{2})=0 \implies a^2=1+2k, \quad k \in \Bbb N$$
and in this case the limit is $0$.
In the book the suggested solution is $a^2=1+2k$ for which the limit is $$ \frac{(-1)^k \cdot (2k+1)^2\cdot\pi}{24}$$
but I don't understand this solution.
Trying to solve the limit with $a^2=1+2k$:
$$ \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} \sim \frac{-\sin (\pi \cdot \frac{1-\cos ax}{x^2}) \cdot \pi \cdot  \frac{\sin (ax)  a  x^2 - (1-\cos (ax))  2x}{x^4}}{2x} $$$$ =  \sin \left( \frac{\pi}{2}+k \pi\right) \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4}$$$$= (-1)^k \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4} $$

Although a solution is provided (and was accepted) I wanted provide a solution using Maclaurin series of cos:
$$
cos(x)=1-x^2/2!-x^4/4!+x^6/6!-...
\\ cos(ax) = 1-a^2x^2/2!-a^4x^4/4!+a^6x^6/6!-...
\\ 1-cos(ax) = a^2x^2/2!+a^4x^4/4!-a^6x^6/6!+...
\\ \frac{(1-cos(ax)}{x^2} = a^2/2!+a^4x^2/4!-a^6x^4/6!+... 
\\ \lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} = \lim_{x\to 0} \frac{\cos \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)}{x^2} 
$$
Since the denominator in $\lim_{x\to 0}$ goes to zero, in order for the limit exist, the numerator also should go to zero for $\lim_{x\to 0}$:
$$
\lim_{x\to 0} {\cos \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)} = \lim_{x\to 0} {\cos \left(\pi \cdot a^2/2 \right)}
$$
for ${\cos \left(\pi \cdot a^2/2 \right)}$ to be zero $a^2$ should be $1+2k$
In this case we can apply the L'Hospital's Rule:
$$
\lim_{x\to 0} \frac{\cos \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)}{x^2} = \lim_{x\to 0} \frac{-\sin \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)(\pi (a^4(2x)/4!-4a^6x^3/6!+... ))}{2x}
= {-\sin \left(\pi \cdot a^2/2 \right)\pi a^4/4!} = \pm \pi a^4/24
$$
for which it is positive if k is even, and it is negative if k is odd.
