I saw a solution to the problem
$$ \lim_{\alpha \to 0} \int_{\alpha}^{1+\alpha} \frac{dx}{1 + x^2 + \alpha^2}$$
that involved simply by moving the limit inside the integral which simplifies to an elementary integral that further gives the limit of $\frac\pi 4$. That is:
$$\lim_{\alpha \to 0} \int_{\alpha}^{1+\alpha} \frac{dx}{1 + x^2 + \alpha^2} = \int_{\alpha}^{1+\alpha} \lim_{\alpha \to 0} \frac{dx}{1 + x^2 + \alpha^2} = \int_{0}^{1} \frac{dx}{1 + x^2} = \arctan1 - \arctan0 = \dfrac \pi 4 $$
There are no remarks explaining why it is allowed to move the limit in this case. As far as I know, the monotone convergence theorem and the dominated convergence theorem allow us to do this when we have infinite limits, as explained in the answers here. In this case, by what theorem is it allowed?
2 things -
you need a change of variables $y=x-\alpha$ to move the $\alpha$ out of the integration limits. In particular $\int_\alpha^{1+\alpha} \lim_{\alpha\to 0}$ is nonsense. An alternative is to write the integration limits as multiplication with an indicator $\mathbb 1_{[\alpha,1+\alpha]}$
To apply the usual sequential DCT/MCT, use the sequential formulation of the limit. Let $\alpha_n \to 0$ be arbitrary. Then what you need is $\lim_{n\to\infty} \int_0^1 \frac{dy}{1+(y+\alpha_n)^2 +\alpha_n^2}$ which is amenable to textbook versions of the theorems