Q: Trouble understanding $\lim_{z \to 0} \frac{1}{z} = \infty$ in complex 
Is it true, that $$\lim_{z \to 0} \frac{1}{z} = \infty$$ for
  $z \in \mathbb{C}$?

I have some trouble with a definition of this. I know that in $\mathbb{C}$ we have only one point at infinity unlike in $\mathbb{R}$ where we have $\pm \infty$. Could somebody provide me with a definition of the above limit?

A: The limit $\lim_{z\to 0}\frac1z =\infty$ means that for all $B>0$, there exists a $\delta>0$ such that $$\left|\frac{1}{z}\right|>B$$whenever $0<|z|<\delta$.
With $\delta=1/B$, we are done.

A: If
$z = x+iy$,
$\frac1{z}
=\frac1{x+iy}
=\frac{x-iy}{x^2+y^2}
=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}
$.
If
$y = \sqrt{x}$,
then,
as $x \to 0$,
$\frac1{z}
=\frac{x}{x^2+x}-i\frac{\sqrt{x}}{x^2+x}
\approx 1-i\frac{1}{\sqrt{x}}
$,
so the real part is bounded
while the imaginary part blows up.
Similarly,
if $x = \sqrt{y}$,
 the imaginary part is bounded
while the real part blows up.
In all cases,
the magnitude of
$\frac1{z}$
blows up as
$|z| \to 0$.
