Prove (by definition) that $\lim _{z\to 1+i}\left(\frac{1}{z^2+1}\right)=\frac{1}{2i+1}$ How can I take this expression $$\left(\frac{1}{z^2+1}\right)-\frac{1}{2i+1}$$ to the form $z-(1+i)$?
I must use the definition of limit to prove this result. That is, use $\delta$ and $\epsilon$.
What I have done so far is to raise the definition: $$\lim _{z\to 1+i}\left(\frac{1}{z^2+1}\right)=\frac{1}{2i+1}\Leftrightarrow[\forall\epsilon>0,\exists\delta>0/\forall z\in\mathbb{C}:(0<|z-(1+i)|<\delta\Rightarrow|\left(\frac{1}{z^2+1}\right)-\frac{1}{2i+1}|<\epsilon)]$$
I must prove:
$$\left|\left(\frac{1}{z^2+1}\right)-\frac{1}{2i+1}\right|<\epsilon$$ as long as $0<|z-(1+i)|<\delta$
We start from
\begin{align}
\left|\left(\frac{1}{z^2+1}\right)-\frac{1}{2i+1}\right|&=\left|\frac{1+2i-(z^2+1)}{(z^2+1)(2i+1)}\right|\\
&=\left|\frac{2i-z^2}{(z^2+1)(2i+1)}\right|\\
&=\left|\frac{(2i-z^2)(1-2i)}{(z^2+1)(2i+1)(1-2i)}\right|\\
&=\left|\frac{2i+4+i(2z^2+2)}{5(z^2+1)}\right|.
\end{align}
But I still don't get the expression I need.
 A: It is unfortunate that most textbooks and teachers do not teach how to use asymptotic analysis. This is an intuitive and powerful technique that can be used even if you want to obtain an ε-δ proof.
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
\def\abs#1{\left|#1\right|}
$
First let's use the technique:
  As $t\overset{∈ℂ}→0$:
    $\lfrac1{(1+i+t)^2+1} = \lfrac1{2i+2·(1+i)·t+t^2+1}$.
    $⊆ \lfrac1{2i+1+O(t)} ⊆ \lfrac1{2i+1}·\lfrac1{1+O(t)} ⊆ \lfrac1{2i+1}·(1+O(t)) → \lfrac1{2i+1}$.
That's it!
Now how to convert it into an ε-δ proof? Easy. Simply use concrete bounds in place of the asymptotic bounds:
  Let $r = \abs{t}$.
  Then $\lfrac1{(1+i+t)^2+1} = \lfrac1{2i+1}·\lfrac1{1+u}$   where $u = 2·\lfrac{1+i}{2i+1}·t+\lfrac1{2i+1}·t^2$.
  And $\abs{u} ≤ 2·r+r^2 ≤ 3r$   if $r ≤ 1$.
  And $\abs{\lfrac1{1+u}-1} = \lfrac{\abs{u}}{\abs{1+u}} ≤ \lfrac{\abs{u}}{1-\lfrac12} ≤ 6r$   if $r ≤ 1$ and $\abs{u}≤\lfrac12$.
  Thus $\abs{ \lfrac1{2i+1}·(\lfrac1{1+u}-1) } ≤ \lfrac12·6r = 3r$   if $r ≤ 1$ and $3r ≤ \lfrac12$.
  Thus $\abs{ \lfrac1{(1+i+t)^2+1} - \lfrac1{2i+1} } ≤ 3r$   if $r ≤ \lfrac16$.
It is now trivial to turn this into an ε-δ proof.
