Question

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

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

Answer 1

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

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