Use l’Hôpital’s Rule to find the limits in Exercises 85–108.
99. lim(x→0) (2^sin(x) - 1)/(e^x - 1)

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First, recognize that the limit is of the form \( \frac{0}{0} \) when \( x \to 0 \), since \( 2^{\sin(0)} - 1 = 2^0 - 1 = 0 \) and \( e^0 - 1 = 0 \). This means l'Hôpital's Rule can be applied.

Apply l'Hôpital's Rule by differentiating the numerator and denominator separately with respect to \( x \).

Differentiate the numerator: \( \frac{d}{dx} \left( 2^{\sin(x)} - 1 \right) = \frac{d}{dx} \left( 2^{\sin(x)} \right) = 2^{\sin(x)} \cdot \ln(2) \cdot \cos(x) \) using the chain rule.

Differentiate the denominator: \( \frac{d}{dx} (e^x - 1) = e^x \).

Rewrite the limit using these derivatives: \[ \lim_{x \to 0} \frac{2^{\sin(x)} \cdot \ln(2) \cdot \cos(x)}{e^x} \]. Then, evaluate this new limit by substituting \( x = 0 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

l’Hôpital’s Rule

l’Hôpital’s Rule is a method for evaluating limits that result in indeterminate forms like 0/0 or ∞/∞. It states that the limit of a ratio of functions can be found by taking the limit of the ratio of their derivatives, provided certain conditions are met.

Limits and Indeterminate Forms

Understanding limits involves analyzing the behavior of functions as the input approaches a specific value. Indeterminate forms such as 0/0 occur when direct substitution does not yield a clear limit, signaling the need for alternative techniques like l’Hôpital’s Rule.

Derivatives of Exponential and Trigonometric Functions

Calculating the derivatives of functions like 2^sin(x) and e^x is essential for applying l’Hôpital’s Rule. This involves using the chain rule and the fact that the derivative of a^u is a^u ln(a) times the derivative of u, and the derivative of e^x is e^x.

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