Limits of Real and Imaginary Parts

Theorem
Let $f: D \to \C$ be a complex function, where $D \subseteq \C$.

Let $z_o \in D$ be a complex number.

Suppose $f$ is continuous at $z_0$.

Then:


 * $\displaystyle \lim_{z \to z_o} \operatorname{Re} \left({f \left({z}\right) }\right) = \operatorname{Re} \left({ \lim_{z \to z_o} f \left({z}\right) }\right)$


 * $\displaystyle \lim_{z \to z_o} \operatorname{Im} \left({f \left({z}\right) }\right) = \operatorname{Im} \left({ \lim_{z \to z_o} f \left({z}\right) }\right)$

Here, $\operatorname{Re} \left({f \left({z}\right) }\right) $ denotes the real part of $f \left({z}\right)$, and $\operatorname{Im} \left({f \left({z}\right) }\right) $ denotes the imaginary part of $f \left({z}\right)$.

Proof
By definition of continuity:


 * $\forall \epsilon > 0: \exists \delta > 0: \left\vert{z - z_0}\right\vert < \delta \implies \left\vert{f \left({z}\right) - f \left({z_0}\right)}\right\vert < \epsilon$

Given $\epsilon > 0$, we find $\delta > 0$ so for all $z \in \C$ with $\left\vert{z - z_0}\right\vert < \delta$:

It follows that:


 * $\forall \epsilon > 0: \exists \delta > 0: \left\vert{z - z_0}\right\vert < \delta \implies \left\vert{ \operatorname{Re} \left({ f \left({z}\right) }\right) - \operatorname{Re} \left({f \left({z_0}\right)} \right)} \right\vert < \epsilon$

Then:

The proof for imaginary parts follows when $\operatorname{Re}$ is replaced by $\operatorname{Im}$ in the equations.