If $$\lim\limits_{n\to\infty}[(\cos\frac{k\pi}{4})^n-(\cos\frac{k\pi}{6})^n] = 0$$ then prove that either $k$ is divisible by $24$ or $k$ is divisible neither by $4$ nor by $6$
How to prove this
If $\lim\limits_{n\to\infty}[(\cos\frac{k\pi}{4})^n-(\cos\frac{k\pi}{6})^n] = 0$ then k is
2
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limits
1 Answers
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Let $k=12m$ where $m\in\mathbb{Z}$ then $$\lim\limits_{n\to\infty}[(\cos\frac{k\pi}{4})^n-(\cos\frac{k\pi}{6})^n] = 0$$ $$\lim\limits_{n\to\infty}[(\cos(3m\pi)^n-(\cos(2m\pi)^n] = 0$$ $$\lim\limits_{n\to\infty}[(-1)^{3mn}-(-1)^{2mn}] = 0$$ $$\lim\limits_{n\to\infty}[(-1)^{3mn}-1] = 0$$ iff $2\ell=3mn$ or $8\ell=kn$for a $\ell\in\mathbb{Z}$. This valid iff $\dfrac{k}{12}\in\mathbb{Z}$ and $\dfrac{k}{8}\in\mathbb{Z}$. This proves that either $k$ is divisible by $24$ or $k$ is divisible neither by $4$ nor by $6$.