Use l’Hôpital’s rule to find the limits in Exercises 7–52.
32. lim (x → 0) (3^x - 1) / (2^x - 1)

Verified step by step guidance

1. First, recognize that as \(x \to 0\), both the numerator \(3^x - 1\) and the denominator \(2^x - 1\) approach 0, creating an indeterminate form \(\frac{0}{0}\). This is a perfect situation to apply l'Hôpital's Rule.
2. Recall l'Hôpital's Rule states that if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then the limit can be found by evaluating \(\lim_{x \to a} \frac{f'(x)}{g'(x)}\), provided this latter limit exists.
3. Differentiate the numerator and denominator separately with respect to \(x\). For the numerator, \(f(x) = 3^x - 1\), use the derivative formula for exponential functions: \(\frac{d}{dx} a^x = a^x \ln(a)\). So, \(f'(x) = 3^x \ln(3)\).
4. Similarly, differentiate the denominator \(g(x) = 2^x - 1\) to get \(g'(x) = 2^x \ln(2)\).
5. Now, apply l'Hôpital's Rule by taking the limit of the ratio of derivatives as \(x \to 0\): \(\lim_{x \to 0} \frac{3^x \ln(3)}{2^x \ln(2)}\). Since \(3^0 = 1\) and \(2^0 = 1\), substitute these values to simplify the limit expression.

Key Concepts

Limits and Indeterminate Forms

Limits describe the behavior of a function as the input approaches a particular value. When direct substitution results in an indeterminate form like 0/0 or ∞/∞, special techniques such as l’Hôpital’s rule are needed to evaluate the limit.

l’Hôpital’s Rule

l’Hôpital’s rule states that if a limit yields an indeterminate form 0/0 or ∞/∞, the limit of the ratio of the functions equals the limit of the ratio of their derivatives, provided this latter limit exists. It simplifies evaluating tricky limits by differentiating numerator and denominator separately.

Derivatives of Exponential Functions

The derivative of an exponential function a^x with respect to x is a^x times the natural logarithm of a (ln a). This rule is essential when applying l’Hôpital’s rule to limits involving exponential expressions like 3^x and 2^x.