Step 1: Limit Evaluation Setup We must evaluate the following limit:\[L = \lim_{x \to 0} \frac{35^x - 7^x - 5^x + 1}{(e^x - e^{-x}) \ln(1 - 3x)}.\]
Step 2: Numerator Approximation Analysis Using the first-order approximation \( a^x - 1 \approx x \ln a \) for small \( x \), we have:\[35^x - 1 \approx x \ln 35, \quad 7^x - 1 \approx x \ln 7, \quad 5^x - 1 \approx x \ln 5.\]The numerator can be expressed as:\[35^x - 7^x - 5^x + 1 \approx x \ln 35 - x \ln 7 - x \ln 5 = x (\ln 35 - \ln 7 - \ln 5).\]Since \( \ln 35 = \ln (7 \times 5) = \ln 7 + \ln 5 \), the expression becomes:\[\ln 35 - \ln 7 - \ln 5 = (\ln 7 + \ln 5) - \ln 7 - \ln 5 = 0.\]This indicates that the numerator's behavior for small \( x \) is of order \( x^2 \), necessitating a second-order approximation.
Step 3: Denominator Approximation For small \( x \), the approximations are:\[e^x - e^{-x} \approx 2x.\]And \( \ln(1 - 3x) \approx -3x \). The denominator thus simplifies to:\[% Option(2x)(-3x) = -6x^2.\]
Step 4: Limit Computation The second-order approximation for the numerator is found to be:\[(1 - 7^x)(1 - 5^x) \approx (\ln 7 \cdot x)(\ln 5 \cdot x) = x^2 \ln 7 \ln 5.\]Substituting these approximations into the limit expression:\[L = \lim_{x \to 0} \frac{x^2 \ln 7 \ln 5}{-6x^2}.\]After canceling \( x^2 \), we get:\[L = \frac{\ln 7 \ln 5}{-6}.\]
Step 5: Option Verification The computed limit is \( \frac{\ln 5 \cdot \ln 7}{-6} \). This result corresponds to option (D).Final Answer: The limit evaluates to (D) \( \frac{\ln(5) \cdot \ln(7)}{-6} \).