<s>[INST] $\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 $a^2=1+2k$:
$$ \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} \sim \frac{-\sin (\pi \cdot \frac{1-\cos ax}{x^2}) \cdot \pi \cdot  \frac{\sin (ax)  a  x^2 - (1-\cos (ax))  2x}{x^4}}{2x} $$$$ =  \sin \left( \frac{\pi}{2}+k \pi\right) \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4}$$$$= (-1)^k \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4} $$
 [/INST] This part of what you wrote is incorrect:
$$\frac{\cos (\pi \cdot \frac{1-\cos (ax)}{x^2})}{x^2}  \sim   \frac{\cos (\pi \cdot \frac{a^2}{2})}{x^2}$$
You're taking the limit of the top of the fraction, but not the bottom. Instead, you need to consider under what circumstances the limit of the entire fraction might exist. Of course, this can only happen if the top tends towards zero, i.e. if
$$\cos (\pi \cdot \frac{a^2}{2}) = 0$$
So the limit might exist if $a^2 = 1 + 2k$ for some $k \in \mathbb N$. Next, assuming $a$ is one of these values, compute the limit of the entire fraction using L'Hospital's rule:
$$
\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} = \lim_{x\to 0} \frac{\frac{d}{dx} \space \cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{\frac{d}{dx} x^2} $$
$$ = \lim_{x\to 0} \frac{-\pi \left(\frac{a\sin(ax)x - 2(1 - \cos(ax))}{x^3}\right)\sin(\pi\cdot\frac{1 - \cos(ax)}{x^2})}{2x}
$$
$$
= \left( \lim_{x\to 0} \frac{-\pi\cdot(a \sin(ax) x - 2\cdot(1 - \cos(ax)))}{2x^4} \right) \left( \lim_{x\to 0} \sin\left(\pi \frac{1 - \cos(ax)}{x^2}\right) \right)
$$
You already know the second limit is $\sin\left(\pi\cdot\frac{a^2}{2}\right) = \pm 1$ (depending on the particular value of $a$). The first limit is straightforward. Just apply L'Hospital's Rule repeatedly until you have your answer.
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