Finding $\lim \limits_{z\to i} \frac{1}{(z-i)^2}$ rigorously I want to find the limit of the following:
$$\lim \limits_{z\to i} \frac{1}{(z-i)^2}$$
And to me, I can see that the denominator is clearly $0$, and since we are in the extended complex plane, I feel I can now immediately say that:
$$\lim \limits_{z\to i} \frac{1}{(z-i)^2}=\infty$$
But this doesn't feel very rigorous or justified. How do I do it in a justified way?
 A: Let
$z = i+w$.
The fraction is then
$\frac1{w^2}$.
For any real $v > 0$,
if
$|w| < 1/\sqrt{v}$,
then
$|\frac1{w^2}|
> v$.
So the limit is $\infty$,
since for any $v > 0$
we can find a neighborhood
of $i$
where the function is
always greater than
$v$.
A: View the extended complex plane as the Riemann sphere, where $ \infty $ is viewed as the north pole of the sphere. To make your argument rigorous, you can use the stereographic projection to see that the limit of the expression is the north pole.
Alternatively, viewing the extended complex plane as a one-point compactification of the complex plane, you simply have to ensure that for any compact subset $ K $ of $ \mathbb{C} $, there exists a punctured neighborhood $ U $ of $ i $ such that $ \dfrac{1}{(z - i)^{2}} \notin K $ for all $ z \in U $.
A: I think this theorem is useful. 
$\displaystyle \lim_{z\to z_{0}}f(z)=\infty$ iff $\displaystyle \lim_{z\to z_{0}}\frac{1}{f(z)}=0$. ($\lim_{z\to w}f(z)=\infty$ if given $\epsilon>0$, there is a $\displaystyle \delta>0$ such that $|z-w|<\delta\implies |f(z)|>\epsilon$).
A: When evaluating $\lim_{x\to x_0}f(x)$ for a real function, a requirement for the limit to exist, is that the right-hand limit and the left-hand limit are equal; $\lim_{x\to x_0^+}f(x)=\lim_{x\to x_0^-}f(x)$ then we say that the limit exists and $\lim_{x\to x_0}f(x)=f(x_0)$.
In the complex plan, however, there are not just two but an infinite number of directions to go from $z$ to $z_0$, though only two linearly independent ones (along the real and imaginary axises, respectively). In order for you to rigorously show that the limit $\lim_{z\to z_0}f(z)$ exists, you have to show that it is independent of the direction of approach. 
But as mentioned earlier, the limit does not exist i $\mathbb{C}$, but in $\mathbb{C}^*=\mathbb{C}\cup\{\check{\infty}\}$, you can however show that indeed the limit does exist and is equal to $\check{\infty}$ by using the method mentioned above.
