Q. If $l_{1} = \lim_{x \rightarrow 2^{+}}{(x + \lbrack x\rbrack)},\ \ \ l_{2} = \lim_{x \rightarrow 2^{-}}{(2x - \lbrack x\rbrack)}$ and $l_{3} = \lim_{x \rightarrow \pi/2}\frac{\cos x}{(x - \pi/2)}$, then
A) $l_{1} < l_{2} < l_{3}$
B) $l_{2} < l_{3} < l_{1}$
C) $l_{3} < l_{2} < l_{1}$
D) $l_{1} < l_{3} < l_{2}$
Explanation
(c)
$l_{1} = \lim_{x \rightarrow 2^{+}}\left( x + \lbrack x\rbrack \right)$
$= \lim_{h \rightarrow 0}2 + h + \lbrack 2 + h\rbrack = 4$
$l_{2} = \lim_{x \rightarrow 2^{-}}{(2x - \lbrack x\rbrack)}$
$= \lim_{h \rightarrow 0}\{ 2(2 - h) - \lbrack 2 - h\rbrack\}$
$= \lim_{h \rightarrow 0}\{ 2(2 - h) - 1\} = 3$
$l_{3} = \lim_{x \rightarrow \frac{\pi}{2}}\frac{\cos x}{x - \frac{\pi}{2}} = \lim_{x \rightarrow \frac{\pi}{2}} - \sin x = - 1$
[by L’Hospital’s rule]
Thus, $l_{3} < l_{2} < l_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.