Question

Let $f\left(x\right)=\frac{{\sin }^{-1}(1-\left\{x\right\}){\cos }^{-1}(1-\left\{x\right\})}{\sqrt{2\left\{x\right\}}(1-\left\{x\right\})}$,

where $\left\{\right\}$ denotes fractional part of $x,$ then:

• A. ${lim}_{x\to 0^{+}}f\left(x\right)$ is equal to $\frac{\pi }{4}$
• B. ${lim}_{x\to 0^{+}}f\left(x\right)$ is equal to $\sqrt{2}$ times ${lim}_{x\to 0^{-}}f\left(x\right)$
• C. ${lim}_{x\to 0^{-}}f\left(x\right)$ is equal to $\frac{\pi }{2\sqrt{2}}$
• D. ${lim}_{x\to 0^{-}}f\left(x\right)$ is equal to $\frac{\pi }{4\sqrt{2}}$

Answer 1

Answer: (B), (C)

The correct options are
B ${lim}_{x\to 0^{+}}f\left(x\right)$ is equal to $\sqrt{2}$ times ${lim}_{x\to 0^{-}}f\left(x\right)$
C ${lim}_{x\to 0^{-}}f\left(x\right)$ is equal to $\frac{\pi }{2\sqrt{2}}$

${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0}\frac{{\sin }^{-1}(1-x){\cos }^{-1}(1-x)}{\sqrt{2x}(1-x)}$

Put ${\cos }^{-1}(1-x)=\theta$

Then ${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{\theta \to 0}\frac{\frac{\pi }{2}\theta -{\theta }^2}{\sqrt{2(1-\cos \theta )}\cos \theta }$

$={lim}_{\theta \to 0}\frac{\theta (\frac{\pi }{2}-\theta )}{2\left|\sin \frac{\theta }{2}\right|\cos \theta }=\frac{\pi }{2}$
$\Rightarrow {lim}_{x\to 0^{+}}f\left(x\right)=\frac{\pi }{2}$

$={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0}\frac{{\sin }^{-1}(-x){\cos }^{-1}(-x)}{\sqrt{2(1+x)}(-x)}$

$={lim}_{x\to 0}\frac{{\sin }^{-1}x(\pi -{\cos }^{-1}x)}{\sqrt{2(1+x)}x}=\frac{\pi }{2\sqrt{2}}$

$\Rightarrow {lim}_{x\to 0^{-}}f\left(x\right)=\frac{\pi }{2\sqrt{2}}$