Question

lf $l_1=\underset{x\to 2^{-}}{lim}(x+[x\left]\right),l_2=\underset{x\to 2^{-}}{lim}(2x-[x\left]\right)$ and
$l_3=\underset{x\to \frac{\pi }{2}}{lim}\frac{\cos x}{x-\frac{\pi }{2}}$ then:

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

Answer 1

The correct option is C $l_3<l_2=l_1$

$l_1=(2-h)+[2-h](whereh\to 0)$
$=3$
$l_2=2(2-h)-[2-h]$
$=3$
$l_3=\begin{array}{c}lim\\ x\to 0\end{array}\frac{\cos (\frac{\pi }{2}-x)}{-x}$
$=\frac{\sin x}{-x}=-1$