Question

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

• A. -1
• B. 0
• C. 1
• D. $\infty$

Answer 1

The correct option is C 1

$y=\frac{1}{2}si{n}^{-1}\left(\frac{2yx}{x^2+y^2}\right)$

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

$Now{lim}_{y\to 0}1=\frac{1}{x}$

$x=1$