Assuming that the sequence of complex numbers $\{z_n\}$ has the limit $\lim_{n\to\infty}z_n = z_0$, I'm asked to prove that the limit $$ \lim_{n\to\infty} \frac{1}{z_n - z_0} \tag{1}\label{eq:limit} $$ does not exist. By definition, assuming that $\lim_{n\to\infty} z_n = z_0$ is the same as assuming that $ |z_n - z_0| < \varepsilon $ for all $n$ greater than some natural $N(\varepsilon)$ given any positive real $\varepsilon$. If this is true, than it follows that $$ \frac{1}{|z_n - z_0|} > \frac{1}{\varepsilon} $$ for all $n>N(\varepsilon)$, but if $\varepsilon$ is taken arbitrarily small, then $1/\varepsilon$ is arbitrarily large, that is, the real sequence $1/|z_n - z_0|$ is bounded from below, but not from above and, therefore, the limit \eqref{eq:limit} doesn't exist.
Is this reasoning correct? How could I make it more rigorous?
You can proceed as follows. By hypothesis, $$\{z_n\} \to z_0 \Leftrightarrow \forall \varepsilon > 0, \exists N \in \mathbb N \mbox{ such that } \forall n \geq N \ |z_n - z_0| < \varepsilon$$ Note that this implies that $ \left|\frac{1}{z_n- z_0} \right|> \frac{1}{\varepsilon}$. Using proof by contradiction, assume the statement holds true. $$\left\{\frac{1}{z_n- z_0}\right\} \to L \Leftrightarrow \forall \varepsilon > 0, \exists N' \in \mathbb N \mbox{ such that } \forall n \geq N' \ \left|\frac{1}{z_n- z_0} - L\right| < \varepsilon$$ Using the triangular inequality $$\left|\frac{1}{z_n- z_0} - L\right| \geq \left|\frac{1}{z_n- z_0} \right| - |L| $$ Therefore $$ \left|\frac{1}{z_n- z_0} \right| < |L| + \varepsilon $$ By the hypothesis, $$\frac{1}{\varepsilon} <\left|\frac{1}{z_n- z_0} \right| < |L| + \varepsilon \Rightarrow \frac{1}{\varepsilon} < |L| + \varepsilon \leq \varepsilon$$ $$\frac{1}{\varepsilon} \leq \varepsilon \Leftrightarrow \varepsilon \geq 1$$ Which is indeed a contradiction since we assumed that $\varepsilon > 0$ arbitrary.