Show that $\int_{-1}^{1} \frac{1}{\sqrt{1-x^2}} \ dx$ exists. (trouble understanding solutions) The question comes from Spivak's Calculus, Chapter 14 problem 28(e).
Preliminary steps were made to split the integral into the form
$$
\int_{-1}^{1}\frac{1}{\sqrt{1-x^2}} \ dx = 2 \cdot \lim_{\epsilon \to -1^+}\int_{\epsilon}^{0}\frac{1}{\sqrt{1-x^2}} \ dx
$$
However, I do not quite understand the concluding statement made in the solutions, which states that:
... by part(a), $\lim_{\epsilon \to -1^+}\int_{\epsilon}^{0}\frac{1}{\sqrt{1+x}} \ dx$ exists. It can be shown that $\frac{1}{\sqrt{1+x}} \gt \frac{1}{\sqrt{1-x^2}} $ for $-1 < x < 0$, then it follows that $\lim_{\epsilon \to -1^+}\int_{\epsilon}^{0}\frac{1}{\sqrt{1-x^2}} \ dx$ exists as well.

In this case clearly $\lim_{x \to -1^+}\frac{1}{\sqrt{1+x}} = \infty$ and $\lim_{x \to -1^+}\frac{1}{\sqrt{1-x^2}} = \infty$. I have two questions with regards to this:

*

*How does it follow that because $\lim_{\epsilon \to -1^+}\int_{\epsilon}^{0}\frac{1}{\sqrt{1+x}} \ dx$ exists, then $\lim_{\epsilon \to -1^+}\frac{1}{\sqrt{1-x^2}} \ dx$ exists as well?


*Most importantly, is this a generalizable concept? (i.e. if the limits of two functions $f, g \rightarrow \infty$ as $x \rightarrow a$,  and $f > g$ in some interval, then if $\lim_{x \to a} \int_{b}^{x} f$ exists, $\lim_{x \to a}\int_{b}^{x} g$ exists as well.)

This is because, if you denote $\:F(x)=\int_b^xf(t)\,\mathrm dt$, $\:G(x)=\int_b^x g(t)\,\mathrm dt$, $F(x)$ and $G(x)$ are increasing functions of $x$ when $x$ is close to $a$, and
$$G(x) \le F(x)\le \int_b^a\!\!f(t)\,\mathrm dt,$$
i.e. $G(x)$ is increasing and bounded.
