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

To solve \( \lim _{x \rightarrow 0} \frac{2 \sin x - 2 x}{3 x^{2}} \), we can simplify the expression. Notice that we can factor out 2: \[ \lim _{x \rightarrow 0} \frac{2(\sin x - x)}{3 x^{2}} = \frac{2}{3} \lim _{x \rightarrow 0} \frac{\sin x - x}{x^{2}}. \] Now we can use L'Hôpital's Rule since both the numerator and the denominator approach 0 as \( x \) approaches 0. Differentiating the numerator and denominator gives: Numerator's derivative: \( \cos x - 1 \). Denominator's derivative: \( 2x \). Now we have: \[ \lim _{x \rightarrow 0} \frac{\cos x - 1}{2x}. \] Applying L'Hôpital’s Rule again gives us: \[ \lim _{x \rightarrow 0} \frac{-\sin x}{2} = \frac{-0}{2} = 0. \] Thus, the limit simplifies to: \[ \frac{2}{3} \cdot 0 = 0. \] Therefore, \[ \lim _{x \rightarrow 0} \frac{2 \sin x - 2 x}{3 x^{2}} = 0. \]