Question

Given a real-valued function f such that
$f(x)=\begin{cases}\frac{\tan ^{2}\{x\}}{\left(x^{2}-[x]^{2}\right)} & \text { for } x>0 \\ 1 & \text { for } x=0 \\ \sqrt{\{x\} \cot \{x\}} & \text { for } x<0\end{cases}$
where $[x]$ is the integral part and $\{x\}$ is the fractional part of $x$, then

• A. $\displaystyle\lim _{x \rightarrow 0^{+}} f(x)=1$
• B. $\displaystyle\lim _{x \rightarrow 0^{-}} f(x)=\cot 1$
• C. $\cot ^{-1}\left(\displaystyle\lim _{x \rightarrow 0^{-}} f(x)\right)^{2}=1$
• D. $\tan ^{-1}\left(\displaystyle\lim _{x \rightarrow 0^{+}} f(x)\right)=\frac{\pi}{4}$

Answer 1

Answer: (A), (B), (C), (D)

We have, $\displaystyle \lim _{x \rightarrow 0^{+}} f(x)=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{\tan ^{2}\{x\}}{\left(x^{2}-[x]^{2}\right)}$
$=\displaystyle\lim _{x \rightarrow 0^{+}} \frac{\tan ^{2} x}{x^{2}}=1\,\,\, ...(i)$
$\left(\because x \rightarrow 0^{+} ;[ x ]=0 \Rightarrow \{ x \}- x \right)$
Also, $\displaystyle\lim _{ k \rightarrow 0^{-}} f ( x )=\displaystyle\lim _{ x \rightarrow 0^{-}} \sqrt{\{ x \} \cot \{ x \}}=\sqrt{\cot 1}$
$\left(\because x \rightarrow 0^{-} ;[ x ]=-1 \Rightarrow \{ x \}= x +1 \Rightarrow \{ x \} \rightarrow 1\right)$
Also, $\cot ^{-1}\left(\displaystyle\lim _{x \rightarrow 0^{-}} f(x)\right)^{2}=\cot ^{-1}(\cot 1)=1$