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!


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


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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