Prove: limit of complex function exits $\iff$ real part and imaginary part converges The definition of the limit: 
$$\lim_{z\to z_0}f(z)=L$$
If for all $\epsilon>0$ there is $\delta>0$ s.t for all $z$,  $0<|z-z_0|<\delta\Rightarrow |f(z)-L|<\epsilon$
How do we prove that for $L=A+iB$ we need to show that:
$\lim_{z\to z_0}\operatorname{Re}(z)=A$ and $\lim_{z\to z_0}\operatorname{Im}(z)=B$?
I have started looking at$$|f(z)-L|=|x+iy-(A+iB)|=|x-A+i(y-B)|=\sqrt{(x-A)^2+(y-B)^2}<\epsilon$$
Using triangle inequality we get $$\sqrt{(x-A)^2+(y-B)^2}<\sqrt{(x-A)^2}+\sqrt{(y-B)^2}=|x-A|+|y-B|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
Which is $\lim_{z\to z_0}\operatorname{Re}(z)=A$ and $\lim_{z\to z_0}\operatorname{Im}(z)=B$?
 A: If $\lim_{z\to z_0}f(z)=A+Bi$, you want to prove that $\lim_{z\to z_0}\operatorname{Re}\bigl(f(z)\bigr)=A$. Take $\varepsilon>0$. Then there is a $\delta>0$ such that $|z-z_0|<\delta\Longrightarrow\bigl|f(z)-(A+Bi)\bigr|<\varepsilon$. But, if $|z-z_0|<\delta$,$$\varepsilon>\bigl|f(z)-(A+Bi)\bigr|=\sqrt{\bigl(\operatorname{Re}\bigl(f(z)\bigr)-A\bigr)^2+\bigl(\operatorname{Im}\bigl(f(z)\bigr)-B\bigr)^2}\geqslant\bigl|\operatorname{Re}\bigl(f(z)\bigr)-A\bigr|.$$ This proves that $|z-z_0|<\delta\Longrightarrow\bigl|\operatorname{Re}\bigl(f(z)\bigr)-A\bigr|<\varepsilon$. Therefore, $\lim_{z\to z_0}\operatorname{Re}\bigl(f(z)\bigr)=A$ and, by a similar argument, $\lim_{z\to z_0}\operatorname{Im}\bigl(f(z)\bigr)=B$.
If $\lim_{z\to z_0}\operatorname{Re}\bigl(f(z)\bigr)=A$ and $\lim_{z\to z_0}\operatorname{Im}\bigl(f(z)\bigr)=B$, take $\varepsilon>0$. There is a $\delta>0$ such that $|z-z_0|<\delta\Longrightarrow\bigl|\operatorname{Re}\bigl(f(z)\bigr)-A\bigr|<\frac\varepsilon{\sqrt2}$ and $\bigl|\operatorname{Im}\bigl(f(z)\bigr)-B\bigr|<\frac\varepsilon{\sqrt2}$. So\begin{align*}\bigl|f(z)-(A+Bi)\bigr|&=\sqrt{\bigl(\operatorname{Re}\bigl(f(z)\bigr)-A\bigr)^2+\bigl(\operatorname{Im}\bigl(f(z)\bigr)-B\bigr)^2}\\&\leqslant\sqrt{\frac{\varepsilon^2}2+\frac{\varepsilon^2}2}\\&=\varepsilon.\end{align*}
