Prove by definition that $\lim\limits_{z \to i} \dfrac{4z+i}{z+1} = \dfrac{5i}{i+1}$ I'm having trouble to prove this limit using $\epsilon - \delta$.
I know that, for every $\epsilon > 0$, exists $\delta > 0$, such that
$0 < |z-i|<\delta \implies \left|\dfrac{4z+i}{z+1} - \dfrac{5i}{1+i}\right| < \epsilon$
we have
$\left|\dfrac{4z+i}{z+1} - \dfrac{5i}{1+i}\right| = \left|\dfrac{4(z-i)-iz-1}{(z+1)(1+i)} \right|$
Now we can multiply by $i/i$
$\hspace{3.5cm} = \left|\dfrac{4i(z-i)+z-i}{i(z+1)(1+i)} \right|$
$\hspace{3.5cm} = \left|\dfrac{(z-i)(4i+1)}{i(z+1)(1+i)} \right| $
and I don't know what to do from here, I was trying to make appear the term $(z-i)$ to find $\delta$ in terms of $\epsilon$, but I got stuck. Is this the right way to solve this problem? any tips will be helpful.
 A: If $|z-i|<\frac{1}{4},$ show that $|1+z|>1.$ So ensure $\delta \leq\frac{1}4.$
To prove this, you can use the inequality:
$$\left||u|-|v|\right|\leq |u-v|$$ using $u=1+i,v=i-z.$
Also use that $|4i+1|=\sqrt{17}<5,|1+i|=\sqrt 2>1.$
Then $$\left|\frac{(z-i)(4i+1)}{(1+z)(1+i)}\right|\leq |z-i|\frac{\sqrt{17}}{\sqrt 2}<5|z-i|$$
So let $\delta=\min(1/4,\epsilon/5).$
A: $$\left|\frac{(z-i)(4i+1)}{i(z+1)(1+i)} \right|=\left|\frac{(z-i)}{(z+1)} \right|\frac{|4i+1|}{|i|\cdot|1+i|}<3\left|\frac{z-i}{z+1} \right|$$
$|z+1|$ represents the distance of $z$ from $-1$, where $z\in N_\delta(i)$. We will find a non-zero lower bound to it. Suppose $\delta\le1$, then the minimum distance of $z$ from $-1$ is obtained when $z$ is the intersection point of the circle $|\omega-i|=\delta$ and the line joining $-1$ and centre at $i$.

That is$$\begin{align*}|z+1|&\ge\Big|[(0,i)-\delta(\cos\pi/4,i\sin\pi/4)]-(-1,0)\Big|\\
&=\sqrt{2\left(1-\frac\delta{\sqrt2}\right)^2}\\
&\ge\sqrt2\left(1-1/\sqrt2\right)\\&>1/3\end{align*}$$
giving us $|z+1|^{-1}\le3$. Thus$$3\frac{|z-i|}{|z+1|}<9\delta$$ so we may select $\delta=\epsilon/9$. What happens if for a particular $\epsilon,\delta=\epsilon/9>1$? Well, if $\delta>1$ is valid, surely $\delta=1$ is valid as well. Thus $\delta=\min\{\epsilon/9,1\}.\blacksquare$
