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You have provided an image showing the limit expression:

\[
\lim_{{x \to 0}} \frac{x}{{e^x - 1}}
\]

To solve this, we'll evaluate the limit as \( x \to 0 \).

### Step-by-step solution:

1. **Substitute \( x = 0 \) into the expression**:
   \[
   \frac{0}{{e^0 - 1}} = \frac{0}{1 - 1} = \frac{0}{0}
   \]
   This gives an indeterminate form \( \frac{0}{0} \), which means we can apply **L'Hôpital's Rule**.

2. **Apply L'Hôpital's Rule**: 
   L'Hôpital's Rule states that if the limit of the form \( \frac{0}{0} \) exists, we can differentiate the numerator and denominator and take the limit of the resulting expression.
   
   - Differentiate the numerator: \( \frac{d}{dx}(x) = 1 \).
   - Differentiate the denominator: \( \frac{d}{dx}(e^x - 1) = e^x \).

   The limit becomes:
   \[
   \lim_{{x \to 0}} \frac{1}{e^x} = \frac{1}{e^0} = 1
   \]

Thus, the value of the limit is \( 1 \).

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Evaluate the limit: lim (x -> 0) [ x / (e^x - 1) ]

Learn how to solve the limit problem lim (x -> 0) [ x / (e^x - 1) ] by applying L'Hôpital's Rule. This solution covers differentiation of both the numerator and denominator to find the limit as x approaches 0. Suitable for undergraduate calculus students.

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