$\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 $a^2=1+2k$:
$$ \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} \sim \frac{-\sin (\pi \cdot \frac{1-\cos ax}{x^2}) \cdot \pi \cdot  \frac{\sin (ax)  a  x^2 - (1-\cos (ax))  2x}{x^4}}{2x} $$$$ =  \sin \left( \frac{\pi}{2}+k \pi\right) \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4}$$$$= (-1)^k \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4} $$

You need to ensure that $$\dfrac{1-\cos ax} {x^2}\to\dfrac{2k+1}{2}$$ so that the limit in question exists. This clearly means that $a\neq 0$ and further $$a^2=2k+1$$ and thus $k\geq 0$.
Now we can write expression in question as $$(-1)^k\cdot\dfrac {\sin\left(\dfrac{(2k+1)\pi}{2}\left(1-\dfrac{2}{2k+1}\cdot \dfrac{1-\cos ax}{x^2}\right)\right)}{x^2}$$ which can be written as $(-1)^{k}(\sin f(x)) /x^2$ where $f(x) \to 0$. And we rewrite it as $$(-1)^k\cdot \frac{\sin f(x)}{f(x)} \cdot\frac{f(x)} {x^2}$$ Thus the desired limit equals the limit of $$(-1)^k\cdot\frac{1}{x^2}\cdot\frac{(2k+1)\pi}{2}\left(1-\frac{2}{2k+1}\cdot\frac{1-\cos ax} {x^2}\right)=(-1)^k\cdot \frac{\pi} {2}\frac{a^2x^2-4\sin^2(ax/2)}{x^4}$$ Putting $ax=2t$ we can see that the above equals $$(-1)^k\frac{a^{4}\pi}{8}\frac{t^2-\sin^2t}{t^4}$$ You can easily check via factoring that last factor tends to $1/3$ and hence the desired limit is $$(-1)^k\cdot\frac{(2k+1)^2\pi}{24}$$
