<s>[INST] How can I find the limit of this fraction? ($0/0$ type) Suppose I have two complex numbers $g_0$ and $g_1$, when $g_0\rightarrow1$ and $g_1\rightarrow0$, how can I evaluate
$$
\sqrt\frac{g_1^2}{1-g_0^2}
$$
Intuitively I think that should be infinity, but I'm not pretty sure how can I justify that? Thanks!!
 [/INST] Let
$g_1 = a$
and
$g_0 = 1-b$
so
$\sqrt\frac{g_1^2}{1-g_0^2}
=\sqrt\frac{a^2}{1-(1-b)^2}
=\sqrt\frac{a^2}{b(2-b)}
$
where
$a, b \to 0$.
If
$a=b$ this is
$\sqrt\frac{b^2}{b(2-b)}
=\sqrt\frac{b}{(2-b)}
\to 0
$.
If $a = b^{1/4}$ this is
$\sqrt\frac{b^{1/2}}{b(2-b)}
=\sqrt\frac{1}{b^{1/2}(2-b)}
\to \infty
$.
If $a = cb^{1/2}$ this is
$\sqrt\frac{c^2b}{b(2-b)}
=\sqrt\frac{c^2}{(2-b)}
\to \dfrac{c^2}{2}
$
so this can approach any positive value.
</s>