<s>[INST] Does $\lim_{\epsilon \to 0^+}(\int_{1/2}^{1-ε}＋\int_{1＋ε}^{3/2})\frac {\log x} {(x-1)^2} dx$ exist? Does $$\lim_{\epsilon \to 0^+}  \left(\int_{1/2}^{1-ε}＋\int_{1＋ε}^{3/2}\right)\frac{\log x}{(x-1)^2}  dx$$ exist?
Hint says $$\lim_{\epsilon \to 0^+} \left( \frac{\log x-a-b(x-1)-c(x-1)^2}{(x-1)^3}\right)$$ exists only if $a=0,b=1,c=-\frac12$.
I have no ideal how to apply this hint.
Another way is also appreciated. Thank you in advance.
 [/INST] Look at the Taylor expansion of $\log x$ around $x = 1$:
$$\log x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots.$$
Therefore, there exists a continuous function $g : [\frac{1}{2}, \frac{3}{2}] \to \mathbb{R}$ defined by
$$g(x) = -\frac{1}{2} + \frac{x-1}{3} - \frac{(x-1)^2}{4} + \cdots$$
such that $\log x = (x-1) + (x-1)^2 g(x)$.
Therefore,
$$\lim_{\varepsilon\to 0^+} \left( \int_{1/2}^{1-\varepsilon} + \int_{1+\varepsilon}^{3/2} \right) \frac{\log x}{(x-1)^2} dx =
\lim_{\varepsilon\to 0^+} \left( \int_{1/2}^{1-\varepsilon} + \int_{1+\varepsilon}^{3/2} \right) \left( \frac{1}{x-1} + g(x) \right) dx.$$
Now, if we split into the term for $\frac{1}{x-1}$ and the term for $g(x)$, in the first we get
$$\lim_{\varepsilon\to 0^+} (\log(1/2) - \log(\varepsilon)) + (\log(\varepsilon) - \log(1/2)) = \lim_{\varepsilon\to 0^+} 0 = 0.$$
In the second, setting $G$ to be an antiderivative of $g$ on $[1/2, 3/2]$ (and thus $G$ must in particular be continuous) and using the Fundamental Theorem of Calculus, we get
\begin{align*}
\lim_{\varepsilon\to 0^+} (G(3/2) - G(1 + \varepsilon)) + (G(1 - \varepsilon) - G(1/2)) = \\ (G(3/2) - G(1)) + (G(1) - G(1/2)) = G(3/2) - G(1/2).
\end{align*}
Since both components of the sum have a limit, the sum must then also have a limit.  (In other words, the Cauchy principal value of the improper integral $\int_{1/2}^{3/2} \frac{\log x}{(x-1)^2} dx$ exists.)
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