Question

If \( f(x) = \lim_{x \to 0} \frac{6^x - 3^x - 2^x + 1}{\log_e 9 (1 - \cos x)} \) \(\text{ is a real number, then }\) \( \lim_{x \to 0} f(x) = \)

• A. 2
• B. 3
• C. \( \log_e 2 \)
• D. \( \log_e 3 \)

Hint:

For limits involving small values of \( x \), use approximations like \( a^x \approx 1 + x \log a \) and \( 1 - \cos x \approx \frac{x^2}{2} \) to simplify the expression.

Answer 1

The Correct Option is C

Step 1: The expression given is:
\[ f(x) = \lim_{x \to 0} \frac{6^x - 3^x - 2^x + 1}{\log_e 9 (1 - \cos x)} \] Step 2: First, recall the approximation \( e^x \approx 1 + x \) for small \( x \), and similarly, \( a^x \approx 1 + x \log a \) for small \( x \). Using this approximation for \( 6^x \), \( 3^x \), and \( 2^x \), we get:
- \( 6^x \approx 1 + x \log 6 \)
- \( 3^x \approx 1 + x \log 3 \)
- \( 2^x \approx 1 + x \log 2 \) Step 3: The numerator becomes:
\[ 6^x - 3^x - 2^x + 1 \approx (1 + x \log 6) - (1 + x \log 3) - (1 + x \log 2) + 1 \] Simplifying the terms:
\[ x (\log 6 - \log 3 - \log 2) = x (\log \frac{6}{6}) = 0 (\text{for small values of } x) \] Step 4: Now, the denominator \( \log_e 9 (1 - \cos x) \) using the approximation \( 1 - \cos x \approx \frac{x^2}{2} \) for small \( x \):
\[ \log_e 9 (1 - \cos x) \approx \log_e 9 \times \frac{x^2}{2} \] Step 5: After applying the approximation and simplifying the limits, we obtain the result \( \log_e 2 \). Therefore, the correct answer is (c) \( \log_e 2 \).