Question

If $\underset{x\to a}{lim}\left[{\sin }^{-1}\frac{2x}{1+x^2}\right]$ doesn't exist, then the number of possible value(s) of $a$ is
(Here, $[.]$ denotes the greatest integer function)

Answer 1

The limit of greatest integer function does not exist at those points where the function attains integral values.
Range of ${\sin }^{-1}$ function is $[-\frac{\pi }{2},\frac{\pi }{2}]$.
So, $\underset{x\to a}{lim}\left[{\sin }^{-1}\frac{2x}{1+x^2}\right]$ doesn't exist at those values of $x$ where ${\sin }^{-1}\frac{2x}{1+x^2}=-1,0,1$

For ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=1$
$\frac{2x}{1+x^2}=\sin 1$
$\Rightarrow \left(\sin 1\right)x^2-2x+\sin 1=0$
$\Rightarrow x=\frac{2\pm \sqrt{4-4{\sin }^{2}1}}{2\sin 1}$

Similarly, for ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=-1,$ we get two distinct values of $x$.
$x=\frac{-2\pm \sqrt{4-4{\sin }^{2}1}}{2\sin 1}$

For ${\sin }^{-1}\frac{2x}{1+x^2}=0$
$\frac{2x}{1+x^2}=0$
$\Rightarrow x=0$

Hence, total number of values of $a$ is $5.$