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If $l_1=\lim _{x \rightarrow 2^{+}}(x+[x]), l_2 \lim _{x \rightarrow 2^{-}}(2 x-[x])$ and $l_3=\lim _{x \rightarrow \pi / 2} \frac{\cos x}{(x-\pi / 2)}$, then :
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Verified Answer
The correct answer is:
$l_3 < l_2 < l_1$
$l_1=\lim _{x \rightarrow 2^{+}} x+[x]$
$=\lim _{h \rightarrow 0} 2+h+[2+h]=4$
$l_2=\lim _{x \rightarrow 2^{-}}(2 x-[x])$
$=\lim _{h \rightarrow 0}\{2(2-h)-[2-h]\}$
$=\lim _{h \rightarrow 0}\{2(2-h)-1\}=4-1=3$
$l_3=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{x-\frac{\pi}{2}}=\lim _{x \rightarrow \frac{\pi}{2}} \sin x=1$
(By L' Hospital's rule)
Thus, $l_3 < l_2 < l_1$
$=\lim _{h \rightarrow 0} 2+h+[2+h]=4$
$l_2=\lim _{x \rightarrow 2^{-}}(2 x-[x])$
$=\lim _{h \rightarrow 0}\{2(2-h)-[2-h]\}$
$=\lim _{h \rightarrow 0}\{2(2-h)-1\}=4-1=3$
$l_3=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{x-\frac{\pi}{2}}=\lim _{x \rightarrow \frac{\pi}{2}} \sin x=1$
(By L' Hospital's rule)
Thus, $l_3 < l_2 < l_1$
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