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!!
 A: 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.
A: I want to say  again that your question is not well defined as it is stated now. We can modify $g_0$ and $g_1$ so that we have infinitely many solutions. If I'm not mistaken, you want to calculate the double limit
$$
\lim_{(g_0,g_1)\to(1,0)}\frac{g_1}{\sqrt{1-g_0^2}}.
$$

Let $S$ be an open set containing $(x_0,y_0)$ , and let $f$  be a function of two variables defined on $S$ , except possibly at  $(x_0,y_0)$ . The limit of  $f(x,y)$  as  $(x,y)$  approaches  $(x_0,y_0)$  is  $L$ , denoted
$$\lim_{(x,y)\to (x_0,y_0)}f(x,y)=L,$$
means that given any $\varepsilon>0$, there exists $\delta>0$  such that for all  $(x,y)≠(x_0,y_0)$ , if $(x,y)$  is in the open disk centered at $(x_0,y_0)$  with radius $\delta$ , then  $|f(x,y)−L|<\varepsilon$.

To show our limit is $0$, let $\varepsilon$ be given. We want to find $\delta>0$ such that if $\sqrt{(g_0-1)^2+(g_1-0)^2}<\delta$, then $$\left|\frac{g_1}{\sqrt{1-g_0^2}}-0\right|<\varepsilon.$$
Observe that $g_1<\delta$. Let $1/\sqrt{1-g_0^2}$ be bounded by a finite constant $\kappa$. (Don't forget that $g_0\neq1$ for this proof.) Choose $\delta<\varepsilon/\kappa$.
$$
\begin{align}
\left|\frac{g_1}{\sqrt{1-g_0^2}}\right|<{\delta}{\kappa}<\frac{\varepsilon\kappa}{\kappa}=\varepsilon.
\end{align}
$$
And this concludes the proof
$$\lim_{(g_0,g_1)\to(1,0)}\frac{g_1}{\sqrt{1-g_0^2}}=0.$$

Addendum: Looking at another interpretation of your limit would be way easier:
$$
\lim_{g_0\to1}\lim_{g_1\to0}\frac{g_1}{\sqrt{1-g_0^2}}=\lim_{g_0\to1}\frac{0}{\sqrt{1-g_0^2}}=0.
$$
It is also equal to zero but it is not the same as the first limit we calculated above!
