If l1=Ltx→2+x+x, l2=Ltx→2-2x-x and l3=Ltx→π/2cos xx-π2 then

I1=ltx→2+x+x=ltx→2+x+2=2+2=4I2=ltx→2−2x−x=ltx→2−2x−1=22−1=4−1=3I3=ltx→π2cosxx−π2=ltx→π2sinπ2−x−π2−x=−1I3<I2<I1

If l1=Ltx→2+x+x, l2=Ltx→2-2x-x and l3=Ltx→π/2cos xx-π2 then

I1=ltx→2+x+x=ltx→2+x+2=2+2=4I2=ltx→2−2x−x=ltx→2−2x−1=22−1=4−1=3I3=ltx→π2cosxx−π2=ltx→π2sinπ2−x−π2−x=−1I3<I2<I1

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$If l_{1} = \underset{x \to 2^{+}}{Lt} (x + [x]), l_{2} = \underset{x \to 2^{-}}{Lt} (2 x - [x]) and l_{3} = \underset{x \to π / 2}{Lt} \frac{\cos x}{(x - \frac{π}{2})} then$

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$l_{1} < l_{2} < l_{3}$

$l_{2} < l_{3} < l_{1}$

$l_{3} < l_{2} < l_{1}$

$l_{1} < l_{3} < l_{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} I_{1} = \underset{x \to 2^{+}}{l t} x + [x] \\ = \underset{x \to 2^{+}}{l t} x + 2 = 2 + 2 = 4 \\ I_{2} = \underset{x \to 2^{-}}{l t} 2 x - [x] \\ = \underset{x \to 2^{-}}{l t} 2 x - 1 \\ = 2 (2) - 1 = 4 - 1 = 3 \\ I_{3} = \underset{x \to \frac{π}{2}}{l t} \frac{\cos x}{x - \frac{π}{2}} \\ = \underset{x \to \frac{π}{2}}{l t} \frac{\sin (\frac{π}{2} - x)}{- (\frac{π}{2} - x)} \\ = - 1 \\ I_{3} < I_{2} < I_{1} \end{array}$