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Suppose that $f:(0,\infty)\rightarrow \mathbb{R}$ is a function such that $f(x)=\lim_{n\rightarrow \infty}\cos^n (\frac{1}{n^x})$. If $|\cos(1/n^x)|<1$, then $f(x)=0$. We want to find a number $x$ such that $\cos (1/n^x)=\pm 1$, so that $f(x)=\pm 1$. At such $x$, $f$ would be discontinuous. How to locate such point in $(0,\infty).$ We may find a sequence $\{x_n\}$ which is convergent in $(0,\infty)$ and for which $\lim_{n\rightarrow}\cos ^n(1/n^{x_n})=1.$

Nanasaheb Phatangare
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  • Given the question history, I could not find any improvement in the quality of your questions. Please improve the question by providing your own ideas and/or contexts. Also, check this posting to learn how to ask a good question. – Sangchul Lee Sep 14 '19 at 14:26
  • It seems to find $x$ such that $1/n^x=m\pi$ for $n\in \mathbb{N}$. – Nanasaheb Phatangare Sep 14 '19 at 14:26
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    You can get some insights from here. – rtybase Sep 14 '19 at 14:37
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    Using the fact that $\log\cos(t)\sim-t^2/$ as $t\to 0$, meaning that the ratio of two functions converges to $1$ as $t\to0$, we get $$\log\left(\cos^n(n^{-x})\right)\sim n \cdot \left(-\frac{n^{-2x}}{2}\right)=-\frac{n^{1-2x}}{2}.$$ Using this, you can easily conclude that $$f(x)=\begin{cases}0,&x<\frac{1}{2}\e^{-1/2},&x=\frac{1}{2}\1,&x>\frac{1}{2}\end{cases}$$ – Sangchul Lee Sep 14 '19 at 14:42

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