Let R be such that the function [f(x)=cases^-1(1-x^2)^-1(1-x)x-x^3, & x0, & x=0cases] is c…

Question

Let $\alpha\in\mathbb{R}$ be such that the function
$$f(x)=\begin{cases}\dfrac{\cos^{-1}(1-\{x\}^2)\sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x\ne0\\alpha, & x=0\end{cases}$$
is continuous at $x=0$, where $\{x\}=x-[x]$, $[x]$ is the greatest integer $\le x$. Then:

Accepted Answer

As $x	o0^-$, $\{x\}	o1$; as $x	o0^+$, $\{x\}	o0$.

LHL $=\displaystyle\lim_{x	o0}\frac{\cos^{-1}(-x^2-2x)}{(1+x)(2+x)}=\frac{\cos^{-1}(0)}{1\cdot2}=\frac\pi4$

RHL $=\displaystyle\frac\pi2\lim_{x	o0}\frac{\cos^{-1}(1-x^2)}{x}=\frac\pi2\cdot2\lim_{x	o0}\frac{x}{2x^2-x^4}=\frac{\pi}{\sqrt2}$ (via L'Hopital)

Since LHL $=\dfrac\pi4
e$ RHL $=\dfrac{\pi}{\sqrt2}$, the function is discontinuous at $x=0$ for every $\alpha$. So no such $\alpha$ exists.

Solutions · 1