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Given the sequence

\begin{equation} z_n=\frac{1}{2n\pi}, \quad n \in \mathbb{N} \end{equation} try to evaluate the following limit: \begin{equation} \lim_{z \to z_n} f(z) \end{equation} where $f(z)$ is a function in the complex field, defined as: \begin{equation} f(z)=\left( \frac{1}{z-1} \right) \cos \left( \frac{1}{z} \right) \end{equation}

What I did (amongst other), was to try to evaluate the limit: \begin{equation} \lim_{n\to \infty}\frac{1}{\frac{1}{2n\pi}-1}\cos(2n\pi) \end{equation} which I find it to take values within the range $[-1,1]$, but the author says it is equal to infinite.

Now, that is confusing. Is it really infinite? And if yes, how can this be proved?

Thank you!

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Observe that for all $\;n\in\Bbb Z\;,\;\;\cos 2n\pi=1\;$ , so

$$\lim_{n\to\infty}\frac{2n\pi}{1-2n\pi}\cos2n\pi=-1$$

I don't know what author says this limit is $\;\infty\;$ but I think that either he meant other sequence, other function or he is simply wrong.

Now, you wrote you want the limit

$$\lim_{z\to z_n}f(z)=\lim_{z\to z_n}\left(\frac1{z-1}\right)\cos\frac1z=\left(\frac1{\frac1{2n\pi}-1}\right)\cos2n\pi=\frac{2n\pi}{1-2n\pi}$$

I'm not sure why you then take the limit of the last expression, which is a very different limit of the one with $\;n\to\infty\;$, though.

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