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

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Calculate the limit $$ \lim_{xightarrow 0} \frac{x e^{-x}}{3 e^{x}-3} $$.
Evaluate the limit by following steps:
- step0: Evaluate using L'Hopital's rule:
  $$\lim _{xightarrow 0}\left(\frac{xe^{-x}}{3e^{x}-3}ight)$$
- step1: Rewrite the expression:
  $$\lim _{xightarrow 0}\left(\frac{1}{3e^{x}-3}	imes xe^{-x}ight)$$
- step2: Calculate:
  $$\lim _{xightarrow 0}\left(\frac{1}{3}e^{-x}ight)$$
- step3: Calculate:
  $$\lim _{xightarrow 0}\left(\frac{1}{3}	imes 1ight)$$
- step4: Multiply the terms:
  $$\lim _{xightarrow 0}\left(\frac{1}{3}ight)$$
- step5: Calculate:
  $$\frac{1}{3}$$
The limit of $$ \lim _{x ightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}ight) $$ is $$ \frac{1}{3} $$ or approximately 0.333.
The limit is $$ \frac{1}{3} $$ or approximately 0.333.

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