Q105 / 160 · AP EAMCET 2006 · Mathematics

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 :

  • A
    $l_1 < l_2 < l_3$
  • B
    $l_2 < l_3 < l_1$
  • Correct answer
    C
    $l_3 < l_2 < l_1$
  • D
    $l_1 < l_3 < l_2$

Solution

$l_1=\lim _{x \rightarrow 2^{+}} x+[x]$ $=\lim _{h \rightarrow 0} 2+h+[2+h]=4$ $l_2=\lim _{x \rightarrow 2^{-}}(2 x-[x])$ $=\lim _{h \rightarrow 0}\{2(2-h)-[2-h]\}$ $=\lim _{h \rightarrow 0}\{2(2-h)-1\}=4-1=3$ $l_3=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{x-\frac{\pi}{2}}=\lim _{x \rightarrow \frac{\pi}{2}} \sin x=1$ (By L' Hospital's rule) Thus, $l_3 < l_2 < l_1$

Hence, the correct answer is option: 'C'.

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