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

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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&#x27;Hospital&#x27;s Rule, which states that (lim_{{x to a}} frac{f(x)}{g(x)} = lim_{{x to a}} frac{f&#x27;(x)}{g&#x27;(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.

Answer

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&#x27;Hospital&#x27;s Rule, which states that (lim_{{x to a}} frac{f(x)}{g(x)} = lim_{{x to a}} frac{f&#x27;(x)}{g&#x27;(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.

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

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