2) \( \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) \)

To evaluate the limit \( \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) \), we start by substituting \( x = 0 \): \[ \frac{0 \cdot e^{0}}{3 e^{0} - 3} = \frac{0}{3 - 3} = \frac{0}{0}, \] which is an indeterminate form. Therefore, we can apply L'Hôpital's Rule, which is used for limits of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). We first differentiate the numerator and the denominator: 1. **Numerator**: The derivative of \( x e^{-x} \) can be found using the product rule: \[ \frac{d}{dx}(x e^{-x}) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x). \] 2. **Denominator**: The derivative of \( 3e^{x} - 3 \) is: \[ \frac{d}{dx}(3e^{x} - 3) = 3e^{x}. \] Now we apply L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{x e^{-x}}{3 e^{x} - 3} = \lim_{x \to 0} \frac{e^{-x}(1 - x)}{3 e^{x}}. \] Substituting \( x = 0 \) into the new limit gives: \[ \frac{e^{0}(1 - 0)}{3 e^{0}} = \frac{1}{3}. \] Thus, the limit is: \[ \boxed{\frac{1}{3}}. \]