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(D) For $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$: As $x \rightarrow 2^{+}$,$[x] = 2$. Therefore,$l_1 = 2^2 + 2 = 4 + 2 = 6$. For $l_2 = \lim_{x

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$l_3 < l_2 < l_1$

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(D) For $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$: As $x \rightarrow 2^{+}$,$[x] = 2$. Therefore,$l_1 = 2^2 + 2 = 4 + 2 = 6$. For $l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$: As $x \rightarrow 3^{-}$,$[x] = 2$. Therefore,$l_2 = 2(3) - 2 = 6 - 2 = 4$. For $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$: Let $y = x - \frac{\pi}{2}$. As $x \rightarrow \frac{\pi}{2}$,$y \rightarrow 0$. Then $x = y + \frac{\pi}{2}$. $l_3 = \lim_{y \rightarrow 0} \frac{\cos(y + \frac{\pi}{2})}{y} = \lim_{y \rightarrow 0} \frac{-\sin y}{y} = -1$. Comparing the values,we have $l_3 = -1$,$l_2 = 4$,and $l_1 = 6$. Thus,$l_3 < l_2 < l_1$.

Let $[x]$ denote the greatest integer not exceeding $x$. If $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$,$l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$,then:

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