As shown in the figure, when the circle passing through $F_{1}, F_{2}$ is tangent to the line at point $P$, $\angle F_{1} P F_{2}$ is maximized. Let the line intersect the $x$-axis at point $D$. By the power of a point theorem, we have $D P^{2}=D F_{1} \cdot D F_{2}$ $=3 \sqrt{3} \cdot 5 \sqrt{3} \Rightarrow|D P|=3 \sqrt{5}$. Also, $\triangle D F_{1} P \sim \triangle D P F_{2}$ $\Rightarrow \frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{\left|D F_{1}\right|}{|D P|}=\frac{5 \sqrt{3}}{3 \sqrt{5}}=\frac{\sqrt{15}}{3}$.
5. Given the ellipse $\frac{x^{2}}{1}+y^{2}=1$ with left and right foci $F_{1}, F_{2}$, point $P$ is on the line $l: x+2 \sqrt{3} y-4 \sqrt{3}=0$. When $\angle F_{1} P F_{2}$ takes the maximum value,
$\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}$ is