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'll first simplify the expression inside the limit. As \( x \) approaches 0, both the numerator and the denominator approach 0, resulting in an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule, which is used for determining limits of indeterminate forms. 1. Differentiate the numerator and denominator: - The numerator \( x e^{-x} \) can be differentiated using the product rule: \[ \frac{d}{dx}(x e^{-x}) = e^{-x} + x(-e^{-x}) = e^{-x} (1 - x). \] - The denominator \( 3 e^{x} - 3 \) differentiates to: \[ \frac{d}{dx}(3 e^{x} - 3) = 3 e^{x}. \] 2. Now we apply L'Hôpital's Rule: \[ \lim_{x \rightarrow 0} \frac{x e^{-x}}{3 e^x - 3} = \lim_{x \rightarrow 0} \frac{e^{-x}(1 - x)}{3 e^x}. \] 3. Substituting \( x = 0 \): - The numerator becomes \( e^{0}(1 - 0) = 1 \), - The denominator becomes \( 3 e^{0} = 3 \). Thus, the limit evaluates to \[ \frac{1}{3}. \] Therefore, the final result is: \[ \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) = \frac{1}{3}. \]