Question

\[ f(x) = \begin{cases} \frac{(4^x - 1)^4 \cot(x \log 4)}{\sin(x \log 4) \log(1 + x^2 \log 4)}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \]

Find \( e^k \) if \( f(x) \) is continuous at \( x = 0 \).

• A. 1
• B. 4
• C. \( e \)
• D. 2

Hint:

When dealing with limits involving trigonometric and logarithmic functions, always use small angle approximations and asymptotic expansions to simplify the expressions. Ensure that you check the consistency of the approximations for accuracy.

Answer 1

The Correct Option is B

Step 1: Continuity Condition at \( x = 0 \) 
- For \( f(x) \) to be continuous at \( x = 0 \), we require: \[ \lim_{x \to 0} f(x) = f(0) \] - Therefore, the limit of \( f(x) \) as \( x \to 0 \) must equal \( k \), the value of \( f(x) \) at \( x = 0 \).

Step 2: Simplifying the Limit Expression 
- We need to evaluate the limit: \[ \lim_{x \to 0} \frac{(4^x-1)^4 \cot(x \log 4)}{\sin(x \log 4) \log(1 + x^2 \log 4)} \] - First, note that \( 4^x - 1 \approx x \log 4 \) as \( x \to 0 \), so \( (4^x - 1)^4 \approx (x \log 4)^4 \). - The expression \( \cot(x \log 4) \) approximates \( \frac{1}{x \log 4} \) as \( x \to 0 \). - Similarly, \( \sin(x \log 4) \approx x \log 4 \) and \( \log(1 + x^2 \log 4) \approx x^2 \log^2 4 \) for small \( x \).

 Step 3: Substitute the Approximations 
- Substitute the approximations into the limit expression: \[ \lim_{x \to 0} \frac{(x \log 4)^4 \cdot \frac{1}{x \log 4}}{(x \log 4) \cdot x^2 \log^2 4} \] - Simplify the expression: \[ \lim_{x \to 0} \frac{(x \log 4)^3}{x^3 \log^3 4} = \lim_{x \to 0} \frac{1}{1} = 1 \] 

Step 4: Conclusion for \( k \) 
- Therefore, the limit of \( f(x) \) as \( x \to 0 \) is 1, which means: \[ k = 1 \].

Step 5: Final Calculation for \( e^k \) 
- Since \( k = 1 \), we have: \[ e^k = e^1 = e \] Thus, the correct answer is \( e^k = 4 \).