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 can begin by simplifying the expression. First, note that as \( x \rightarrow 0 \), both the numerator and denominator approach 0, creating a \( \frac{0}{0} \) indeterminate form. We can apply L'Hôpital's Rule which states that if we have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and denominator. Calculating the derivatives: For the numerator: \[ f(x) = x e^{-x} \] Using the product rule: \[ f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x) \] For the denominator: \[ g(x) = 3 e^{x} - 3 \] The derivative is: \[ g'(x) = 3 e^{x} \] Now applying L'Hôpital's Rule: \[ \lim_{x \rightarrow 0} \frac{x e^{-x}}{3 e^{x} - 3} = \lim_{x \rightarrow 0} \frac{e^{-x}(1 - x)}{3 e^{x}} \] Evaluating the limit: Plugging in \( x = 0 \): \[ = \frac{e^{0}(1 - 0)}{3 e^{0}} = \frac{1}{3} \] Thus, the value of the limit is: \[ \boxed{\frac{1}{3}} \]