<s>[INST] $\lim\limits_ {n \to \infty} \left( \cos ^{2n} \left(\frac{k \pi}{3}\right)-\cos ^{2n}\left(\frac{k \pi}{5}\right)\right )=0$ For how many values of $k=1,2,3,....200$ $$\lim\limits_ {n \to \infty} \left( \cos ^{2n} \left(\frac{k \pi}{3}\right)-\cos ^{2n}\left(\frac{k \pi}{5}\right)\right )=0$$
My Try: 
Let $A=\cos\left(\frac{k \pi}{3}\right)$ and $B=\cos\left(\frac{k \pi}{5}\right)$
The limit approaches zero only if 
$$A=B=1$$ OR
$$A=B=-1$$  OR
$$A=B=0$$
if $A=B=1$ or $A=B=-1$ $\implies$
$$\frac{k \pi}{3}=2m \pi$$ and
$$\frac{k \pi}{5}=2n \pi$$  
OR
$$\frac{k \pi}{3}=(2m+1) \pi$$ and
$$\frac{k \pi}{5}=(2n+1) \pi$$ 
$\implies$ $k$ is a multiple of $15$  which are $13$ in number.
Finally if $A=B=0$ we have
$$\frac{k \pi}{3}=(2m+1) \frac{\pi}{2}$$ and
$$\frac{k \pi}{5}=(2n+1)\frac{\pi}{2}$$ $\implies$
Any idea of how to find number of $k's$ in this case?
 [/INST] A different approach. It is worth mentioning that 
$$0\leq \cos^2\left(\frac{k\pi}{3}\right)\leq1 \text{ and } 0\leq \cos^2\left(\frac{k\pi}{5}\right)\leq1$$
for $\forall k \in \mathbb{Z}$. As a result ($0\leq a \leq 1 \Rightarrow 0\leq a^n \leq 1, n\in \mathbb{N}$)
$$0\leq \cos^{2n}\left(\frac{k\pi}{3}\right)\leq1 \text{ and } 0\leq \cos^{2n}\left(\frac{k\pi}{5}\right)\leq1$$
It is also known (or can easily be shown) that 
$$\lim\limits_{n \rightarrow \infty}a^n=0, 0\leq a<1 \tag{1}$$ 
As a result
$$\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{3}\right)−\cos^{2n}\left(\frac{k\pi}{5}\right)\right)=0 \tag{2}$$
is true when 


*

*$\color{red}{3 \nmid k \text{ and }5 \nmid k}$, because in this case $0\leq \cos^2\left(\frac{k\pi}{3}\right)< 1 \text{ and } 0\leq \cos^2\left(\frac{k\pi}{5}\right)< 1$. It results from $(1)$ and $$\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{3}\right)−\cos^{2n}\left(\frac{k\pi}{5}\right)\right)=\\
\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{3}\right)\right)−\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{5}\right)\right)=\\
0-0=0$$

*$\color{red}{3 \mid k \text{ and } 5 \mid k \text{ or simply } 15 \mid k}$, because in this case $\cos^2\left(\frac{k\pi}{3}\right)=1 \text{ and } \cos^2\left(\frac{k\pi}{5}\right)=1$ and 
$$\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{3}\right)−\cos^{2n}\left(\frac{k\pi}{5}\right)\right)=\\
\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{3}\right)\right)−\lim\limits_{n\rightarrow \infty}\left(\cos^{2n}\left(\frac{k\pi}{5}\right)\right)=\\
1-1=0$$


In all the other case, the limit is either $1$ or $-1$. Those "other cases" are $\color{red}{3 \mid k \text{ or } 5 \mid k \text{ but } 15 \nmid k}$ or playing with inclusion–exclusion principle $\left \lfloor \frac{200}{3} \right \rfloor + \left \lfloor \frac{200}{5} \right \rfloor - 2\left \lfloor \frac{200}{15} \right \rfloor=80$ of cases when $(2)$ is not true. Or $200-80=120$ cases when $(2)$ is true.
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