I'm reading Ahlfors' complex analysis. In the book, he states that
From $\lim_{x \to a} f(x) = A$ we obtain that \begin{align*}\lim_{x \to a} \Re f(x) = \Re A\\ \lim_{x \to a} \Im f(x) = \Im A\end{align*}
where $x$ and $f(x)$ can be real or complex. Here also assuming that $f(x)$ is defined over some open set.
I wanted to quickly verify this explicitly using the $\varepsilon-\delta$ definition of limit.
Since we know that $\lim_{x \to a} f(x) = A$ then we know that $\forall \varepsilon >0, \exists \delta_{(\varepsilon)}>0$ such that if $0 < \lvert x-a \rvert < \delta \implies \lvert f(x)-A \rvert < \varepsilon$. Noting that $$ \lvert f(x)-A \rvert = \bigr\lvert \overline{f(x)-A } \bigl\rvert = \bigr\lvert \overline{f(x)}-\overline{A } \bigl\lvert $$ we see that $$ \bigr\lvert \Re f(x)+ \Re A \bigr\rvert = \Biggr\lvert \frac{f(x) + \overline{f(x)}}{2} + \frac{A + \overline{A}}{2} \Biggl\rvert \le \frac{\bigr\lvert f(x)- A \bigr\rvert}{2} + \frac{\bigr\lvert \overline{f(x)}- \overline{A} \bigr\rvert}{2} = \lvert f(x)-A \rvert< \varepsilon $$ using the same $\varepsilon$ that we used in $f(x) \to A$. And since we know that for that $\varepsilon$ we can always find a $\delta$ such that... Then this also works for $\lim_{x \to a} \Re f(x) = \Re A$. And for the imaginary part we would to an analogous proceadure with $\Im z = \frac{z -\overline{z}}{2i}$.
Is this proof of what the author meant correct? Or am I glossing over some important parts? Thank you!
There are two sign errors, it should be $$ \bigr\lvert \Re f(x) \color{red}{-} \Re A \bigr\rvert = \Biggr\lvert \frac{f(x) + \overline{f(x)}}{2} \color{red}{-} \frac{A + \overline{A}}{2} \Biggl\rvert \le \frac{\bigr\lvert f(x)- A \bigr\rvert}{2} + \frac{\bigr\lvert \overline{f(x)}- \overline{A} \bigr\rvert}{2} = \lvert f(x)-A \rvert< \varepsilon $$ to make it correct.
Alternatively you can use $$ \Re(w)^2 + \Im(w)^2 = |w|^2 \implies |\Re(w)| \le |w|, |\Im(w)| \le |w| $$ for $w = f(z) - A$, compare How to prove that $-|z| \le \Re (z) \le |z|$ and $-|z| \le \Im (z) \le |z|$?. $$ $$