Step 1: $\left|\left(x^2-5\right)-11\right|<1 \Rightarrow\left|x^2-16\right|<1 \Rightarrow-1<x^2-16<1 \Rightarrow 15<x^2<17 \Rightarrow \sqrt{15}<x<\sqrt{17}$.

Step 2: $|x-4|<\delta \Rightarrow-\delta<x-4<\delta \Rightarrow-\delta+4<x<\delta+4$.
Then $-\delta+4=\sqrt{15} \Rightarrow \delta=4-\sqrt{15} \approx 0.1270$, or $\delta+4=\sqrt{17} \Rightarrow \delta=\sqrt{17}-4 \approx 0.1231$; thus $\delta=\sqrt{17}-4 \approx 0.12$.