Question

If $y=\frac{1}{2}{\sin }^{-1}\left(\frac{2xy}{x^2+y^2}\right)$ and y < x then $\underset{y\to 0}{limx}=$

• A. -1
• B. 0
• C. 1
• D. $\infty$

Answer 1

The correct option is C 1

$y=\frac{1}{2}{\sin }^{-1}\left(\frac{2xy}{x^2+y^2}\right)$

$\Rightarrow$ $2y={\sin }^{-1}\left(\frac{2xy}{x^2+y^2}\right)$

$\Rightarrow$ $\sin 2y=\frac{2xy}{x^2+y^2}$

$\Rightarrow$ $x^2+y^2=\frac{2xy}{\sin 2y}$

$\Rightarrow$ $\underset{y\to 0}{lim}{x}^2+y^2=x\underset{y\to 0}{lim}\frac{2y}{\sin 2y}$

$\Rightarrow$ $x^2+0=x\left(1\right)$ [ Since, $\underset{x\to 0}{lim}=\frac{\sin x}{x}=1$ ]

$\Rightarrow$ $x^2-x=0$

$\Rightarrow$ $x(x-1)=0$

$\Rightarrow$ $x=0,1$

since, $y<x$ then $x\ne 0$

$\therefore$ $x=1$