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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