I'm struggling with interpreting how to evaluate $$\lim_{\epsilon\to 0} \int_{[-M, M]\setminus(-\epsilon, \epsilon)} \frac{1}{x} \,dx$$ (The limit is taken from above).
I'm understanding that $[-M, M]\setminus(-\epsilon, \epsilon)$ means the interval $[-M, M]$ without the interval $(-\epsilon, \epsilon)$. At first I thought maybe we're evaluating the integral from $-M$ to $M$ but excluding the area that gets arbitrarly close to $0$, because there the integral would diverge to infinity. With my interpretation the value would simply be $0$ because of the symmetry of $\frac{1}{x}$ about the origin - and if we integrated from $-\epsilon$ to $-M$ and add the integral from $\epsilon$ to $M$ they would cancel each other out, avoiding the problems at the origin.
I would appreaciate a clarification as I'm not sure at all about my reasoning.
Yes, that's right. You can compute explicitly: \begin{align*} \int_{[-M, M]\setminus(-\epsilon, \epsilon)} \frac{1}{x} \,dx &= \int_{-M}^{-\epsilon} \frac{1}{x}\,dx + \int_{\epsilon}^M\frac{1}{x} \,dx \\ &= \left.\ln(-x)\right|_{-M}^{-\epsilon} + \left.\ln(x)\right|_{\epsilon}^M \\ &= \left(\ln \epsilon - \ln M\right) + \left(\ln M - \ln \epsilon\right) = 0 \end{align*} So the limit of the above as $\epsilon\to 0$ is also $0$.