As shown in the figure, in the coordinate plane, the circle $x ^ { 2 } + y ^ { 2 } = 1$ and the curve $y = \ln ( x + 1 )$ meet at point A in the first quadrant. For point $\mathrm { B } ( 1,0 )$, let H be the foot of the perpendicular from point P on arc AB to the $y$-axis, and let Q be the intersection of segment PH and the curve $y = \ln ( x + 1 )$. Let $\angle \mathrm { POB } = \theta$. If $S ( \theta )$ is the area of triangle OPQ and $L ( \theta )$ is the length of segment HQ, and $\lim _ { \theta \rightarrow + 0 } \frac { S ( \theta ) } { L ( \theta ) } = k$, find the value of $60 k$. (Here, $0 < \theta < \frac { \pi } { 6 }$ and O is the origin.) [4 points]
As shown in the figure, in the coordinate plane, the circle $x ^ { 2 } + y ^ { 2 } = 1$ and the curve $y = \ln ( x + 1 )$ meet at point A in the first quadrant. For point $\mathrm { B } ( 1,0 )$, let H be the foot of the perpendicular from point P on arc AB to the $y$-axis, and let Q be the intersection of segment PH and the curve $y = \ln ( x + 1 )$. Let $\angle \mathrm { POB } = \theta$. If $S ( \theta )$ is the area of triangle OPQ and $L ( \theta )$ is the length of segment HQ, and $\lim _ { \theta \rightarrow + 0 } \frac { S ( \theta ) } { L ( \theta ) } = k$, find the value of $60 k$. (Here, $0 < \theta < \frac { \pi } { 6 }$ and O is the origin.) [4 points]