Limit of $f(z)=\frac 1z$ as $z$ approaches $0$?

I know that the answer is infinity, but doesn’t this contradict the theorem stating that $\displaystyle\lim_{z\to0} \Im(f(z)) = \Im(\displaystyle\lim_{z\to0} f(z))$? I mean, the LHS doesn’t exist because $\Im(f(z))$ approaches $-\infty$ and $\infty$ as $z$ approaches $0$ along the imaginary axis from the positive and negative directions respectively. I know I’m misunderstanding something fundamental, but what is it?

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The key is to remember that $\infty$ is not a complex number! In reality, in complex analysis, when we say that
$$\lim_{z \to z_0} f(z) = \infty,$$
what we really mean is
$$\lim_{z \to z_0} |f(z)| = \infty.$$
So we bring it back to real numbers. And of course, this statement doesn’t exactly make sense, since $\infty$ is not a real number, either. What we really mean is that for all (large) $M>0$, we can find a (small) $\delta > 0$ such that for all $z$ where $|z – z_0| < \delta$, we have $|f(z)| > M$. And this is definitely true for $1/z$ as $z \to 0$.

Now, in real analysis, we can often talk about $-\infty$ and $\infty$, while in complex analysis, we can perhaps imagine $\infty e^{i \theta}$ for $\theta \in [0, 2\pi)$. But, it turns out that one of the best ways to think of large complex numbers is that they are actually somewhat close to each other; if we have complex numbers $z$ that we know are large, say $|z|>M$, then we can look at $1/z$ and note that $\left| \frac{1}{z} \right| < 1/M$. Therefore, we know that (as we defined a limit of complex numbers going to $\infty$)
$$\lim_{z \to z_0} f(z) = \infty$$
if and only if
$$\lim_{z \to z_0} \frac{1}{f(z)} = 0.$$

This notion is formalized with the Riemann sphere. The Riemann sphere allows us to think of meromorphic functions as functions where we can “fill-in” the poles by mapping them to $\infty$, and to think of rational functions as mappings from the Riemann sphere to itself. But note that even the Riemann sphere isn’t “enough” to handle essential singularities.

$\lim_{z\to 0}f(z)=\frac 1z$ does not exist in the real numbers, nor in the complex numbers. There is no infinity in either one. If you use the extended reals: $\mathbb R \cup \infty$ or the Riemann sphere $\mathbb C \cup \infty$ you can certainly say (and prove) $\lim_{z\to 0}f(z)=\frac 1z = \infty$ The imaginary part of $\infty$, like the imaginary part of $0$ is undefined in the Riemann sphere.

Thomas’ comment explains why this theorem doesn’t make much sense if the function takes on the value infinity. I’ll try to make more sense of this. You can define the Riemann sphere, $\hat{\Bbb C}$ as the one point compactification of $\Bbb C$. As a set we can think of this as $\hat{\Bbb C}= \Bbb C \cup \{\infty\}$, where if $\infty \not\in U$ then $U \subset \hat{\Bbb C}$ is open if $U$ is open in the standard topology on $\Bbb C$. If $\infty \in U$, then $U$ is open if $U \cap \Bbb C=K^{c}$, where $K \subset \Bbb C$ is compact. In short, the open sets of $\hat{\Bbb C}$ are the standard open sets of $\Bbb C$, and open sets containing $\infty$ are the complements of compact sets in $\mathbb C$.

So, lets think of $f$ as a map from $\Bbb C$ to $\hat{\Bbb C}$, and lets define $f(0)=\infty$. I claim that $f$ is continuous. We show this by showing that for $U \subset \hat{\Bbb C}$ open, $f^{-1}(U)$ is open in $\Bbb C$. Clearly we only need to consider the case where $\infty \in U$, because the other cases follow easily from the fact that $f$ is continuous away from $0$. For this, just follow the argument Twiceler gives.

Hope that helps.