If l_1 = lim_{x rightarrow 2^{+}} (x + [x]),l | vedclass

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(C) For $l_1 = \lim_{x \rightarrow 2^{+}} (x + [x])$: Let $x = 2 + h$ where $h \rightarrow 0^{+}$. Then $[x] = 2$. So,$l_1 = \lim_{h \rightarrow 0} (2

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$l_3 < l_2 < l_1$

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(C) For $l_1 = \lim_{x ightarrow 2^{+}} (x + [x])$: Let $x = 2 + h$ where $h ightarrow 0^{+}$. Then $[x] = 2$. So,$l_1 = \lim_{h ightarrow 0} (2 + h + 2) = 4$.For $l_2 = \lim_{x ightarrow 2^{-}} (2x - [x])$: Let $x = 2 - h$ where $h ightarrow 0^{+}$. Then $[x] = 1$. So,$l_2 = \lim_{h ightarrow 0} (2(2 - h) - 1) = 4 - 1 = 3$.For $l_3 = \lim_{x ightarrow \pi / 2} \frac{\cos x}{x - \pi / 2}$: Using $L'	ext{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$: $\lim_{x ightarrow \pi / 2} \frac{-\sin x}{1} = -\sin(\pi / 2) = -1$.Comparing the values: $l_3 = -1$,$l_2 = 3$,$l_1 = 4$. Thus,$-1 < 3 < 4$,which implies $l_3 < l_2 < l_1$.

If $l_1 = \lim_{x \rightarrow 2^{+}} (x + [x])$,$l_2 = \lim_{x \rightarrow 2^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \pi / 2} \frac{\cos x}{x - \pi / 2}$,then:

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