Question

If \[ y=\frac{1}{2}\sin^{-1}\!\left(\frac{2xy}{x^2+y^2}\right) \] and \(y<x\), then \[ \lim_{y\to 0} x = \]

• A. \(-1\)
• B. \(0\)
• C. \(1\)
• D. \(\infty\)

Hint:

Useful small-angle approximation: \[ \sin^{-1} t \approx t \quad \text{as } t\to 0 \] When limits involve inverse trigonometric functions, first reduce them using standard approximations.

Answer 1

The Correct Option is C

Step 1: As \(y\to 0\), the argument of \(\sin^{-1}\) also tends to \(0\). For small \(t\), \[ \sin^{-1} t \approx t \]
Step 2: Hence, \[ y \approx \frac{1}{2}\left(\frac{2xy}{x^2+y^2}\right) = \frac{xy}{x^2+y^2} \] 
Step 3: Divide both sides by \(y\) (\(y\neq 0\)): \[ 1 \approx \frac{x}{x^2+y^2} \] 
Step 4: Taking the limit \(y\to 0\): \[ 1=\frac{x}{x^2} =\frac{1}{x} \] 
Step 5: Therefore, \[ x=1 \]