Evaluate the limit \(\lim_{{x \to 0}} \frac{{e^x - 1}}{{x}}\)

Step 1 ：Step 1: Notice that the limit is in the indeterminate form \(\frac{0}{0}\) as \(x\) approaches \(0\), so we can use L'Hospital's Rule, which states that \(\lim_{{x \to a}} \frac{f(x)}{g(x)} = \lim_{{x \to a}} \frac{f'(x)}{g'(x)}\) if the limit exists.

Step 2 ：Step 2: Differentiate the numerator and the denominator separately. The derivative of \(e^x\) is \(e^x\), and the derivative of \(x\) is \(1\), so the limit becomes \(\lim_{{x \to 0}} \frac{e^x}{1}\).

Step 3 ：Step 3: Substitute \(x = 0\) into \(\frac{e^x}{1}\) to get the limit.