Question

If $l_1={lim}_{x\to 2^{+}}(x+[x\left]\right),l_2={lim}_{x\to 2^{-}}(2x-[x\left]\right)$ and $l_3$ $={lim}_{x\to \frac{\pi }{2}}\frac{\cos x}{x-\frac{\pi }{2}}$ then [where [ ] denotes G.I.F.]

• A. $l_1<l_2<l_3$
• B. $l_2<l_3<l_1$
• C. $l_3<l_2<l_1$
• D. $l_1<l_3<l_2$

Answer 1

Answer: (C)

$l_3<l_2<l_1$

$l_1=\underset{x\to 2^{+}}{lt}(x+\left[x\right])=4$

$l_2=\underset{x\to 2^{-}}{lt}(2x-\left[x\right])=4-1=3$

$l_3=\underset{x\to \pi /2}{lt}\frac{cosx}{x-\pi /2}$

$=\underset{x-\pi /2\to 0}{lt}\frac{-\sin (x-\pi /2)}{x-\pi /2}=1$

$\therefore$ $l_3<l_2<l_1$ [C]