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I am attempting to prove the following strictly from the definition of limit:

$$\lim\limits_{z \rightarrow 1-i} [x+i(2x+y)] = 1+i$$

In other words, we want to show that $\forall \varepsilon > 0$, $\exists \delta > 0$ such that:

\begin{align} 0 < |(x+iy)-(1-i)| < \delta &\implies |(x+i(2x+y))-(1+i)|<\varepsilon \\\\ &\iff \\\\ 0 < |(x-1)+i(y+1)| < \delta &\implies |(x-1)+i(2x+y-1)|<\varepsilon \end{align}

I observed that we can rewrite the $\varepsilon$-neighborhood as follows:

\begin{align}|(x-1)+i(2x+y-1)| &= |(x-1)+i(2x-2+y+1)| \\ &= |(x-1)+i(2(x-1)+(y+1))| \\ &= |(x-1)(1+2i)+i(y+1)| \\ &< \varepsilon \end{align}

I am not sure if it means anything that the rewritten $\varepsilon$-neighborhood bears some similarity to the $\delta$-neighborhood (that is, the presence of the $1+2i$ term), but in any case, I am unsure of how to proceed.

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  • $\begingroup$ $|x-1| \leq |(x-1)+i(y+1)| < \delta $ $\endgroup$ – Nosrati Mar 4 '17 at 10:32
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Hint:

$|x-1| \le |(x-1)+i(y+1)| < \delta $

$|y+1| \le |(x-1)+i(y+1)| < \delta $

then use triangular inequality with $|(x-1)(1+2i)+i(y+1)|$.

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