$$ \lim _{x \rightarrow 0^{-}} \frac{x-e^{x}+1}{\left(e^{x}-1\right) x} $$ ev-elle dexivddeen 0 ?

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Calculate the limit $$ \lim_{x\rightarrow 0^{-}} \frac{x-e^{x}+1}{(e^{x}-1)x} $$.
Evaluate the limit by following steps:
- step0: Evaluate using L'Hopital's rule:
  $$\lim _{x\rightarrow 0^{-}}\left(\frac{x-e^{x}+1}{\left(e^{x}-1\right)x}\right)$$
- step1: Use the L'Hopital's rule:
  $$\lim _{x\rightarrow 0^{-}}\left(\frac{\frac{d}{dx}\left(x-e^{x}+1\right)}{\frac{d}{dx}\left(\left(e^{x}-1\right)x\right)}\right)$$
- step2: Find the derivative:
  $$\lim _{x\rightarrow 0^{-}}\left(\frac{1-e^{x}}{e^{x}x+e^{x}-1}\right)$$
- step3: Simplify the expression:
  $$\lim _{x\rightarrow 0^{-}}\left(\frac{1-e^{x}}{xe^{x}+e^{x}-1}\right)$$
- step4: Use the L'Hopital's rule:
  $$\lim _{x\rightarrow 0^{-}}\left(\frac{\frac{d}{dx}\left(1-e^{x}\right)}{\frac{d}{dx}\left(xe^{x}+e^{x}-1\right)}\right)$$
- step5: Find the derivative:
  $$\lim _{x\rightarrow 0^{-}}\left(\frac{-e^{x}}{2e^{x}+xe^{x}}\right)$$
- step6: Simplify the expression:
  $$\lim _{x\rightarrow 0^{-}}\left(\frac{-1}{2+x}\right)$$
- step7: Rewrite the expression:
  $$\frac{-1}{\lim _{x\rightarrow 0^{-}}\left(2+x\right)}$$
- step8: Calculate:
  $$\frac{-1}{2}$$
- step9: Rewrite the fraction:
  $$-\frac{1}{2}$$
The limit of the expression $$ \frac{x-e^{x}+1}{(e^{x}-1)x} $$ as $$ x $$ approaches 0 from the left is -0.5.
The limit is -0.5.

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