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Find the following limit:

$$\lim_{x \to 1/2}\frac{\tan(\pi x)}{2x - 1}+\frac{2}{\pi(2x - 1)^2}$$

Using a calculator, I know the function is divergent, but I don't know how to show that it is divergent. I tried to simplify it first, but that just made the function more complicated.

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    $\begingroup$ What does $\lim_{x\to 1/2}=$ mean? $\endgroup$ – Bernard Jan 23 at 22:10
  • $\begingroup$ I am sorry. Just a typo $\endgroup$ – Hung Do Jan 23 at 22:16
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As other did show, it is almost always easier to work around $0$. So $$\lim_{x \to \frac 12}\frac{\tan(\pi x)}{2x - 1}+\frac{2}{\pi(2x - 1)^2}=\lim_{y \to 0}\frac{1-\pi y \cot (\pi y)}{2 \pi y^2}=\pi\lim_{z \to 0}\frac{1-z \cot (z)}{2 z^2}$$

Now, using Taylor expansions $$\frac{1-z \cot (z)}{2 z^2}=\frac {1-z\left(\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}+O\left(z^5\right) \right) } {2z^2}=\frac{1}{6}+\frac{z^2}{90}+O\left(z^4\right)$$ So, your limit is $\frac \pi 6$.

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It seems that the limit exists at $x\to\frac{1}{2}$. Lets' take $x=\frac{1}{2} -\epsilon$; we will set $\epsilon\to0$.

Let's consider $f(x)=\frac{\tan(\pi x)}{2x - 1}+\frac{2}{\pi(2x - 1)^2}$

$$f(\frac{1}{2}-\epsilon)=\frac{\sin(\frac{\pi}{2}-\pi{\epsilon})}{\cos(\frac{\pi}{2}-\pi\epsilon)(-2\epsilon)}+\frac{2}{\pi(2\epsilon)^2}=$$$$\frac{\cos\pi\epsilon}{\sin(\pi\epsilon)(-2\epsilon)}+\frac{1}{2\pi(\epsilon)^2}=\frac{1}{2\pi(\epsilon)^2}-\frac{1-\frac{(\pi\epsilon)^2}{2!}+...}{2\epsilon(\pi\epsilon-\frac{(\pi\epsilon)^3}{3!}+...)}=\frac{1}{2\pi(\epsilon)^2}\left(1-\frac{1-\frac{(\pi\epsilon)^2}{2!}+...}{1-\frac{(\pi\epsilon)^2}{3!}+...}\right)$$

$$\lim_{\epsilon\to0}f(\frac{1}{2}-\epsilon)=\frac{\pi}{6}$$ The same story if we set $x=\frac{1}{2}+\epsilon$ - we get the same limit. So, it seams that both side limits exist.

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