\( \lim _{x \rightarrow 0} (\frac{4^{x}-3^{x}}{3^{x}-2^{x}}) \)

To evaluate the limit \( \lim _{x \rightarrow 0} \left(\frac{4^{x}-3^{x}}{3^{x}-2^{x}}\right) \), we can start by recalling that \( a^x \) approaches 1 as \( x \) approaches 0 for any positive number \( a \). Thus, we can rewrite the terms in the limit: \[ \lim_{x \to 0} \frac{4^x - 3^x}{3^x - 2^x} = \lim_{x \to 0} \frac{(4^x - 1) - (3^x - 1)}{(3^x - 1) - (2^x - 1)} \] Both the numerator and the denominator approach 0 as \( x \to 0 \), allowing us to apply L'Hôpital's Rule. We can differentiate the numerator and denominator with respect to \( x \): 1. For the numerator, \( \frac{d}{dx}(4^x) = 4^x \ln(4) \) and \( \frac{d}{dx}(3^x) = 3^x \ln(3) \). Thus, the derivative of the numerator is: \[ 4^x \ln(4) - 3^x \ln(3) \] 2. For the denominator, \( \frac{d}{dx}(3^x) = 3^x \ln(3) \) and \( \frac{d}{dx}(2^x) = 2^x \ln(2) \). Thus, the derivative of the denominator is: \[ 3^x \ln(3) - 2^x \ln(2) \] Now we apply L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{4^x \ln(4) - 3^x \ln(3)}{3^x \ln(3) - 2^x \ln(2)} \] Substituting \( x = 0 \): \[ = \frac{4^0 \ln(4) - 3^0 \ln(3)}{3^0 \ln(3) - 2^0 \ln(2)} = \frac{\ln(4) - \ln(3)}{\ln(3) - \ln(2)} \] This simplifies to: \[ = \frac{\ln(\frac{4}{3})}{\ln(\frac{3}{2})} \] Thus, the limit is: \[ \boxed{\frac{\ln(4/3)}{\ln(3/2)}} \]