11. Let the area of the closed region bounded by the lines $l _ { 2 } : n x + y - n = 0$, $l _ { 3 } : x + n y - n = 0$ ($n \in \mathbb{N} ^ { * } , n \geq 2$), the $x$-axis, and the $y$-axis be denoted by $S _ { n }$. Then $\lim _ { n \rightarrow \infty } S _ { n } =$ $\_\_\_\_$ $1$. Analysis: $B \left( \frac { n } { n + 1 } , \frac { n } { n + 1 } \right)$, so $OB \perp AC$, $S _ { n } = \frac { 1 } { 2 } \times \sqrt { 2 } \times \frac { n } { n + 1 } \sqrt { 2 } = \frac { n } { n + 1 }$, so $\lim _ { n \rightarrow \infty } S _ { n } = 1$ [Figure]
(5 points) Given that the three vertices of $\triangle ABC$ are $A ( - 1,2 ) , B ( 3,4 ) , C ( 4 , - 2 )$, and point $(x , y)$ is in the interior of $\square ABCD$, then the range of $z = 2 x - 5 y$ is
11. Let the area of the closed region bounded by the lines $l _ { 2 } : n x + y - n = 0$, $l _ { 3 } : x + n y - n = 0$ ($n \in \mathbb{N} ^ { * } , n \geq 2$), the $x$-axis, and the $y$-axis be denoted by $S _ { n }$. Then $\lim _ { n \rightarrow \infty } S _ { n } =$ $\_\_\_\_$ $1$.
Analysis: $B \left( \frac { n } { n + 1 } , \frac { n } { n + 1 } \right)$, so $OB \perp AC$,\\
$S _ { n } = \frac { 1 } { 2 } \times \sqrt { 2 } \times \frac { n } { n + 1 } \sqrt { 2 } = \frac { n } { n + 1 }$, so $\lim _ { n \rightarrow \infty } S _ { n } = 1$\\
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