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

To solve the limit \( \lim_{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) \), we can start by simplifying the expression. First, notice that as \( x \to 0 \), both the numerator and the denominator approach 0, giving us an indeterminate form of type \( \frac{0}{0} \). Thus, we can apply L'Hôpital's rule: 1. Differentiate the numerator and the denominator: - Numerator: \( f(x) = x e^{-x} \) - Using the product rule: \( f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x) \) - Denominator: \( g(x) = 3 e^{x} - 3 \) - The derivative is \( g'(x) = 3 e^{x} \) 2. Now we apply L'Hôpital's rule: \[ \lim_{x \to 0} \frac{x e^{-x}}{3 e^{x} - 3} = \lim_{x \to 0} \frac{e^{-x}(1 - x)}{3 e^{x}} \] 3. Substituting \( x = 0 \): \[ = \frac{e^{0}(1 - 0)}{3 e^{0}} = \frac{1}{3} \] Thus, the final result is: \[ \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) = \frac{1}{3} \]