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I was reading this question which asks to prove the continuity of $$g(\theta) = \frac{1}{2\pi^2\theta^3}-\frac{\pi}{2}\cot(\pi\theta)\csc^2(\pi\theta), 0\le \theta < 1$$ at $\theta=0$

However, by looking at the first term, the function does not appear to be defined at $\theta=0$, and according to wikipedia, a function must be defined at $c$ in order to be continuous at $c$. I am mostly looking for an intuitive explanation, I can't really understand a formal proof since I am not very familiar with infinite series and "big O" notation.

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Ovi
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    Such a thing can be interpreted as an assertion that $\lim\limits_{\theta\to0^{+}}g(\theta)$ exists. And that then the function $g$ can be extended to define $g(0)=\lim\limits_{\theta\to0^{+}}g(\theta)$ (which of course makes it continuous at $0$). – 2'5 9'2 Jan 04 '16 at 07:41

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This is just a lazy way of saying that $g$ (which is a function $(0,1)\to\mathbb{R}$) can be extended to a function $\tilde{g}:[0,1)\to\mathbb{R}$ which is continuous at $0$. That is, there exists some number $c$ such that if you define $$\tilde{g}(\theta)= \begin{cases} g(\theta) & \text{ if $0<\theta<1$} \\ c & \text{ if $\theta=0$} \end{cases}$$ then the function $\tilde{g}$ is continuous at $0$. If you think about what it means for $\tilde{g}$ to be continuous at $0$, it means exactly that $c=\lim_{\theta\to 0^+}g(\theta)$. So such a continuous extension exists iff the limit $\lim_{\theta\to 0^+}g(\theta)$ exists.

Eric Wofsey
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