Let's determine the value of the following number to four decimal places:

$$
2+\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}
$$
Instead of calculating $\sqrt{48}$ to more decimal places, let's try to transform the expression and determine (1) its value. Proceeding from the inner square root outward:

$$
\begin{gathered}
13+\sqrt{48}=12+2 \sqrt{12}+1=(\sqrt{12}+1)^{2} \\
5-(\sqrt{12}+1)=3-2 \sqrt{3}+1=(\sqrt{3}-1)^{2} \\
3+(\sqrt{3}-1)=(3+2 \sqrt{3}+1) / 2=(\sqrt{3}+1)^{2} / 2,
\end{gathered}
$$

thus the value of (1) is equal to that of

$$
A=2+\frac{\sqrt{3}+1}{\sqrt{2}}=2+\frac{\sqrt{2}+\sqrt{6}}{2}
$$

According to the school function table, $7 \sqrt{2}=\sqrt{98}<9.8995$, hence $\sqrt{2}<1.41422$, and $\sqrt{6}<2.4495$. From these,

$$
A<3.93186 \text {. }
$$

We will also show that $3.93185<A$, which is equivalent to the inequality $1.73204 \ldots<\sqrt{3}$ after rearrangement and squaring. Instead, we will prove the slightly sharper inequality $3.4641<2 \sqrt{3}=\sqrt{12}$. According to the table, $0.0641^{2}<0.00411$, so

$$
(3.4+0.0641)^{2}<11.56+0.43588+0.00411=11.99999<12,
$$

as we stated. The value of the expression in question is therefore 3.9319 to four decimal places.