<s>[INST] for which real values of $a$ the limit is finite and different from $0$ $\lim_{ x \to 0 } \frac{ \cos (\pi \cdot \frac{1-\cos ax}{x^2})}{x^2}$ I have to find for which real values of $a$ this limit is finite and different from $0$:
$$\lim_{ x \to 0 }  \frac{ \cos (\pi \frac{1-\cos ax}{x^2})}{x^2}$$
I've tried to change the form of cos but with no success.
According to my book the final result should be for $a^2=2k+1 (k \in N)$ for which $\lim=\frac{(-1)^k (2k+1)^2 \pi}{24}$.
 [/INST] We start from
$$\cos(ax)=1-\frac{a^2x^2}{2}+\frac{a^4x^4}{4!}(1+\epsilon(x))$$
to get
$$\frac{1-\cos(ax)}{x^2}=\frac{a^2}{2}-\frac{a^4x^2}{24}(1+\epsilon(x))$$
If The limit is finite, then necessarily
$$\cos(\pi.\frac{a^2}{2})=0$$
or
$$a^2=1+2k , k\ge 0$$
In this case
$$A=\pi(\frac{a^2}{2}-\frac{a^4x^2}{24}(1+\epsilon(x))=$$
$$\frac{\pi}{2}+k\pi-\frac{a^4x^2\pi}{24}(1+\epsilon(x))$$
and
$$\cos(A)=-(-1)^k\sin(\frac{a^4x^2\pi}{24}(1+\epsilon(x)($$
$$\sim -(-1)^k\frac{a^2x^2\pi}{24}$$
the limit will be
$$-(-1)^k\frac{a^4\pi}{24}=-(-1)^k\frac{(1+2k)^2\pi}{24}$$
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