Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{{\tan }^{2}x}{x^2-{\left[x\right]}^2}& \text{, for}x>0\\ 1& \text{, for}x=0\\ \sqrt{\left\{x\right\}\cot \left\{x\right\}}& \text{, for}x<0\end{array}$ , then which of the following is correct?
(where $\left[x\right]$ is the step up function and $\left\{x\right\}$ is the fractional part function of $x$)

• A. $\underset{x\to 0^{+}}{lim}f\left(x\right)=1$
• B. $\underset{x\to 0^{-}}{lim}f\left(x\right)=1$
• C. ${\cot }^{-1}{\left(\underset{x\to 0^{-}}{lim}f\left(x\right)\right)}^2=1$
• D. none of these

Answer 1

Answer: (A), (C)

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

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

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

Since,${lim}_{x\to 0}\frac{\tan x}{x}=1$ and ${lim}_{x\to 0^{+}}\left[x\right]=0$

Hence, ${lim}_{x\to 0^{+}}f\left(x\right)=1$

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

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

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

$=\sqrt{\cot 1}$

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

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