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

To evaluate the limit \( \lim _{x \rightarrow 0} \frac{\sin ^{2} \frac{x}{3}}{x^{2}} \), we can start by applying the limit properties related to the sine function. First, we use the fact that as \( x \) approaches 0, the expression \( \frac{\sin u}{u} \) approaches 1 when \( u = \frac{x}{3} \). We can rewrite the limit as follows: \[ \lim_{x \rightarrow 0} \frac{\sin^2 \frac{x}{3}}{x^2} = \lim_{x \rightarrow 0} \left( \frac{\sin \frac{x}{3}}{\frac{x}{3}} \cdot \frac{x}{3} \cdot \frac{\sin \frac{x}{3}}{\frac{x}{3}} \cdot \frac{3}{x} \right) \] This simplifies into: \[ = \lim_{x \rightarrow 0} \left( \left( \frac{\sin \frac{x}{3}}{\frac{x}{3}} \right)^2 \cdot \frac{3^2}{x^2} \right) \] By substituting \( u = \frac{x}{3} \), as \( x \to 0 \), \( u \to 0 \). Thus, we change \( x \) in terms of \( u \): \[ x = 3u \quad \text{so} \quad \frac{3}{x} = \frac{1}{u} \] Then rewrite the limit: \[ = \lim_{u \rightarrow 0} \left( \left( \frac{\sin u}{u} \right)^2 \cdot 9 \right) = 9 \cdot \left( \lim_{u \rightarrow 0} \left( \frac{\sin u}{u} \right)^2 \right) \] Since \( \lim_{u \rightarrow 0} \frac{\sin u}{u} = 1 \), we then have: \[ = 9 \cdot 1^2 = 9 \] Thus, the limit is: \[ \boxed{9} \]