Question

Let $f\left(x\right)=\{\begin{array}{c}\frac{{\tan }^2\left\{x\right\}}{x^2-[x{]}^2};x>0\\ 1;x=0\\ \sqrt{\left\{x\right\}\cot \left\{x\right\}};x<0\end{array}$, where $\left\{x\right\}$ denotes fractional part function and $\left[x\right]$ denotes greatest integer function

• A. $\underset{x\to 0^{+}}{lim}f\left(x\right)=1$
• B. $\underset{x\to 0^{-}}{lim}f\left(x\right)=1$
• C. ${\cot }^{-1}\left({lim}_{x\to 0^{-}}f\right(x){)}^2=1$
• D. $f$ is continuous at $x=0$

Answer 1

Answer: (A), (C)

The correct options are
A $\underset{x\to 0^{+}}{lim}f\left(x\right)=1$
C ${\cot }^{-1}\left({lim}_{x\to 0^{-}}f\right(x){)}^2=1$

We have $RHL={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}\frac{{\tan }^2\left\{x\right\}}{(x^2-[x{]}^2)}$
$={lim}_{x\to 0^{+}}\frac{{\tan }^{2}x}{x^2}=1$
$(\because x\to 0^{+},[x]=0\Rightarrow \{x\}=x)$
Also, $LHL={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0}\sqrt{\left\{x\right\}\cot \left\{x\right\}}=\sqrt{\cot 1}$
$(\because x\to 0^{-},[x]=-1\Rightarrow \{x\}=x+1\Rightarrow \{x\}\to 1)$
Here, $LHL\ne RHL$
So, limit of $f\left(x\right)$ does not exist.
Also, ${\cot }^{-1}{\left(\underset{x\to 0^{-}}{lim}f\right(x\left)\right)}^2={\cot }^{-1}\left(\cot 1\right)=1$