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 first analyze the expression as \( x \) approaches 0. First, we observe that both the numerator and the denominator approach 0 as \( x \rightarrow 0 \). This indicates that we can use L'Hôpital's rule, which applies to limits of the form \( \frac{0}{0} \). Now, applying L'Hôpital's rule, we differentiate the numerator and the denominator: 1. The derivative of the numerator \( x e^{-x} \) can be computed using the product rule: \[ \frac{d}{dx}(x e^{-x}) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x) \] 2. The derivative of the denominator \( 3 e^{x} - 3 \) is: \[ \frac{d}{dx}(3 e^{x} - 3) = 3 e^{x} \] Now, we rewrite the limit using these derivatives: \[ \lim_{x \rightarrow 0} \frac{x e^{-x}}{3 e^{x} - 3} = \lim_{x \rightarrow 0} \frac{e^{-x}(1 - x)}{3 e^{x}} \] As \( x \rightarrow 0 \), \( e^{-x} \rightarrow 1 \), \( 1 - x \rightarrow 1 \), and \( 3 e^{x} \rightarrow 3 \): \[ \lim_{x \rightarrow 0} \frac{e^{-x}(1 - x)}{3 e^{x}} = \frac{1 \cdot 1}{3} = \frac{1}{3} \] Thus, the limit evaluates to: \[ \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) = \frac{1}{3} \]