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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.5.32

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

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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.
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.
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, \(f'(x) = 3^x \ln(3)\).
Similarly, differentiate the denominator \(g(x) = 2^x - 1\) to get \(g'(x) = 2^x \ln(2)\).
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.

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