Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. \[ \lim _{x \rightarrow 0} \frac{\sin ^{-1}(x)}{2 x} \] \( \square \)

To find the limit \[ \lim _{x \rightarrow 0} \frac{\sin^{-1}(x)}{2x}, \] we start by substituting \(x = 0\). The sine inverse at zero is zero, making the expression \( \frac{0}{0} \), which is indeterminate. This allows us to use l'Hôpital's Rule. We differentiate the numerator and denominator: 1. The derivative of \(\sin^{-1}(x)\) is \( \frac{1}{\sqrt{1-x^2}} \). 2. The derivative of \(2x\) is \(2\). Applying l'Hôpital's Rule gives: \[ \lim _{x \rightarrow 0} \frac{\sin^{-1}(x)}{2x} = \lim _{x \rightarrow 0} \frac{\frac{1}{\sqrt{1-x^2}}}{2} = \frac{1}{2\sqrt{1-0^2}} = \frac{1}{2}. \] So, \[ \lim _{x \rightarrow 0} \frac{\sin^{-1}(x)}{2x} = \frac{1}{2}. \] Next, you could also approach this using the Taylor series expansion of \(\sin^{-1}(x)\) around \(x = 0\): The series is \( \sin^{-1}(x) = x + \frac{x^3}{6} + O(x^5) \). Thus, when we plug this into our limit: \[ \frac{\sin^{-1}(x)}{2x} = \frac{x + \frac{x^3}{6} + O(x^5)}{2x} = \frac{1}{2} + \frac{x^2}{12} + O(x^4), \] and as \(x \to 0\), the limit approaches \(\frac{1}{2}\).