3) \( \lim _{x \rightarrow 0} \frac{\sin x}{x^{\frac{2}{3}}} \)

To evaluate the limit \( \lim _{x \rightarrow 0} \frac{\sin x}{x^{\frac{2}{3}}} \), we can use the fact that \( \sin x \) behaves like \( x \) as \( x \) approaches 0. More formally, we have the approximation \( \sin x \approx x \) when \( x \) is near 0. Thus, substituting this approximation into our limit gives: \[ \lim _{x \rightarrow 0} \frac{\sin x}{x^{\frac{2}{3}}} \approx \lim _{x \rightarrow 0} \frac{x}{x^{\frac{2}{3}}} = \lim _{x \rightarrow 0} x^{1 - \frac{2}{3}} = \lim _{x \rightarrow 0} x^{\frac{1}{3}}. \] As \( x \) approaches 0, \( x^{\frac{1}{3}} \) also approaches 0. Therefore, we find that: \[ \lim _{x \rightarrow 0} \frac{\sin x}{x^{\frac{2}{3}}} = 0. \]