MCQ
Let $f (x) = \left[ {\begin{array}{*{20}{c}}  {\tfrac{{{{\tan }^2}\,\,\{ \,x\,\} }}{{{x^2}\,\, - \,\,{{[\,x\,]}^2}}}\,\,\,\,} \\   {1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\   {\sqrt {\{ \,x\,\} \,\,\,\cot \,\,\{ \,x\,\} } } \end{array}} \right.\,\,\,\,\,\begin{array}{*{20}{c}}  {for\,\,\,\,\,x\,\, > \,\,0} \\    {for\,\,\,\,\,x\,\, = \,\,0} \\   {for\,\,\,\,\,x\,\, < \,\,0} \end{array}$ where $[ x ]$ is the step up function and $\{ x \}$ is the fractional part function of $x$ , then :
  • A
    $\mathop {Limit}\limits_{x\,\, \to \,\,{0^ + }} f (x) = 1$
  • B
    $\mathop {Limit}\limits_{x\,\, \to \,\,{0^ - }} f (x) = 1$
  • C
    $cot ^{-1} {\left( {\mathop {Limit}\limits_{x\,\, \to \,\,{0^ - }} \,\,\,f\,(x)} \right)^2}= 1$
  • both $(A)$ and $(C)$

Answer

Correct option: D.
both $(A)$ and $(C)$
d
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\tan ^{2} x}{x^{2}-[x]^{2}}$

$\lim _{x \rightarrow 0^{+}} \frac{\tan ^{2} x}{x^{2}\left(1-\frac{[x]^{2}}{x^{2}}\right)}$

$\operatorname{since,\operatorname{lim}_{x\rightarrow0}\frac{\operatorname{tan}x}{x}=1\text{and}\operatorname{lim}_{x\rightarrow0^{+}}[x]=0}$

$\operatorname{Hence,} \lim _{x \rightarrow 0^{+}} f(x)=1$

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \sqrt{\{x\} \cot \{x\}}$

$=\lim _{x \rightarrow 0^{-}} \sqrt{(x-[x]) \cot (x-[x])}$

$=\lim _{x \rightarrow 0} \sqrt{(x+1) \cot (x+1)}$

$=\sqrt{\cot 1}$

$\left(\lim _{x \rightarrow 0^{-}} f(x)\right)^{2}=\cot 1$

$\Rightarrow \cot ^{-1}\left(\lim _{x \rightarrow 0^{-}} f(x)\right)^{2}=1$

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