determining the limit of an integral with a parameter For any $\alpha \in \mathbb{R}$ determine the limit $$\lim_{\substack{> \\ \delta \rightarrow 0}}\int^2_0\frac{1-x}{\delta+x^{\alpha}}dx \qquad \text{in } [-\infty, +\infty].$$
I want to use some kind of convergence theorem, but I'm stuck because $\alpha$ can be any element in $\mathbb{R}$.
 A: Denote for $\alpha \in \mathbb R$ and $\delta \gt 0$
$$f_{\alpha, \delta}(x) = \frac{1-x}{\delta+x^{\alpha}} \text{ and } F_{\alpha}(x) = \frac{1-x}{x^{\alpha}}.$$
For $x \gt 0$, you have
$$\vert f_{\alpha, \delta}(x) \vert \le \vert F_{\alpha}(x) \vert$$ and
$$\lim_{\substack{> \\ \delta \rightarrow 0}} f_{\alpha, \delta}(x) = F_{\alpha}(x).$$
As $\int_0^2 \vert F_{\alpha}(x) \vert \ dx$ is convergent for $\alpha \lt 1$, you can apply Dominated Convergent Theorem - DCT in that case and get
$$\lim_{\substack{> \\ \delta \rightarrow 0}}\int^2_0\frac{1-x}{\delta+x^{\alpha}}dx = \int_0^2 F_{\alpha}(x) \ dx.$$
So let's now suppose that $\alpha \ge 1$.
You'll easily prove that
$$\int_1^2 f_{\alpha, \delta}(x) \ dx$$ is bounded by $3$ for $(\alpha, \delta) \in [1, \infty) \times (0, \infty)$. This implies the implication
$$\lim_{\substack{> \\ \delta \rightarrow 0}}\int_0^1 f_{\alpha, \delta}(x) \ dx = \infty \implies \lim_{\substack{> \\ \delta \rightarrow 0}}\int_0^2 f_{\alpha, \delta}(x) \ dx = \infty.$$
Now for $0 \lt x \le 1$ (and $\alpha \ge 1$ as assumed), we have
$$0 \le \frac{1-x}{\delta + x} \le f_{\alpha, \delta}(x) .$$
We get the desired result as
$$\int_0^1 \frac{1-x}{\delta + x} \ dx = -1+(1+\delta)\ln \left(1 + \frac{1}{\delta}\right)$$ and
$$\lim_{\substack{> \\ \delta \rightarrow 0}}(1+\delta)\ln \left(1 + \frac{1}{\delta}\right) = \infty.$$
Conclusion
$$
\lim_{\substack{> \\ \delta \rightarrow 0}}\int^2_0\frac{1-x}{\delta+x^{\alpha}} \ dx =
\begin{cases}
\infty &\text{for } \alpha \ge 1\\
\int_0^2 \frac{1-x}{x^{\alpha}} \ dx &\text{for } \alpha \lt  1
\end{cases}$$
