I want to know if I answered this correctly:
$$|f(z) - L| = \left|\frac{4z+i}{z+1} - \frac{5i}{1+i}\right|\\ = \left|\frac{4z+i+4zi-1-5zi-5i}{(z+1)(1+i)}\right| \\ =\left| \frac{4z-4i-zi-1}{(z+1)(1+i)}\right|\\ = \left|\frac{4(z-i)-i(z-i)}{(z+1)(1+i)}\right|\\ =\left|\frac{(z-i)(4-i)}{(z+1)(1+i)} \right|$$
Then consider that $|z+1| > r$ and we can assume $\left|\frac{(z-i)(4-i)}{(z+1)(1+i)} \right|\leq \left|\frac{(z-i)(4-i)}{r(1+i)} \right|$. Thus $|z-i|< \frac{\varepsilon r(1+i)}{4-i}= \delta$ . This condition need to be satisfied together with $|z+1| > r$ and for that we can presume $$|z+1| = |z+1-i+i| = |i+1 +(z-i)| \\ \geq |1+i| - |z-i| > \sqrt2 - \delta > \sqrt2 -(\sqrt2 - r) = r.$$ Lastly, $\delta = \min\left\{\frac{{\varepsilon r(1+i)}}{4-i},\sqrt2 - r\right\}$
There is a problem there. Your answer contains a number $r$, but you don't tell us what that number is. Here's a suggestion: take $r=1$. Then take$$\delta=\min\left\{\varepsilon\frac{\sqrt2}{\sqrt{17}},\sqrt2-1\right\}.$$Those numbers $\sqrt2$ and $\sqrt{17}$ are $|1+i|$ and $|4-i|$ respectively.