Calculate the limit \( \lim _{x \rightarrow 0} \frac{x e^{-x}}{3 e^{x}-3} \)

To evaluate the limit \[ L = \lim_{x \rightarrow 0} \frac{x e^{-x}}{3 e^{x}-3}, \] we start by substituting \( x = 0 \): 1. The numerator becomes \( 0 \cdot e^{0} = 0 \). 2. The denominator evaluates to \( 3 e^{0} - 3 = 3 - 3 = 0 \). Since both the numerator and denominator approach 0, we can apply L'Hôpital's Rule, which is appropriate for \( \frac{0}{0} \) indeterminate forms. According to L'Hôpital's Rule, we take the derivative of the numerator and the derivative of the denominator and then take the limit again. First, we differentiate the numerator: \[ \frac{d}{dx}(x e^{-x}) = e^{-x} + x(-e^{-x}) = e^{-x} - x e^{-x} = e^{-x}(1 - x). \] Next, we differentiate the denominator: \[ \frac{d}{dx}(3 e^{x} - 3) = 3 e^{x}. \] Now we can apply L'Hôpital's Rule: \[ L = \lim_{x \to 0} \frac{e^{-x}(1 - x)}{3 e^{x}}. \] Substituting \( x = 0 \) into this expression gives: 1. The numerator becomes \( e^{0}(1 - 0) = 1 \). 2. The denominator becomes \( 3 e^{0} = 3 \). Thus, we have: \[ L = \frac{1}{3}. \] Therefore, the limit is \[ \boxed{\frac{1}{3}}. \]