Question

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

• A. $\underset{x\to 0^{+}}{lim}f\left(x\right)=1$
• B. ${lim}_{x\to 0^{-}}f\left(x\right)=cot1$
• C. $co{t}^{-1}{\left(\underset{x\to 0^{-}}{lim}f\right(x\left)\right)}^2=1$
• D. $\left(\underset{x\to 0^{+}}{lim}f\right(x\left)\right)=\frac{\pi }{4}$

Answer 1

Answer: (A), (C)

$\underset{x\to 0^{+}}{lim}f\left(x\right)=1$
$co{t}^{-1}{\left(\underset{x\to 0^{-}}{lim}f\right(x\left)\right)}^2=1$

We have $\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{ta{n}^2\left\{x\right\}}{(x^2-[x{]}^2)}$
$=\underset{x\to 0^{+}}{lim}\frac{ta{n}^{2}x}{x^2}=1$
$(\because x\to 0^{+},[x]=0\Rightarrow \{x\}=x)$
Also, $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\sqrt{\left\{x\right\}cot\left\{x\right\}}=\sqrt{cot1}$
$(\because x\to 0^{-},[x]=-1\Rightarrow \{x\}=x+1\Rightarrow \{x\}\to 1)$
Also, $co{t}^{-1}{\left(\underset{x\to 0^{-}f\left(x\right)}{lim}\right)}^2=co{t}^{-1}\left(cot1\right)=1$.