Question

If \[ \lim_{x \to 0} \frac{e^{a(x-1)} + 2\cos(bx) + e^{-x}(c - 1)}{x \cos x - \ln(1 + x)} = 2, \] Then the value of \( a^2 + b^2 + c^2 \) is:

• A. 8
• B. 9
• C. 10
• D. 11

Hint:

For limits involving small \( x \), use the Taylor series expansion for the functions involved to simplify the expression and solve for the constants.

Answer 1

The Correct Option is C

Step 1: Use series expansions.
To solve this, we expand the terms \( e^{a(x-1)} \), \( \cos(bx) \), and \( e^{-x}(c-1) \) into their Taylor series around \( x = 0 \). Step 2: Analyze the denominator.
Similarly, expand \( x \cos x - \ln(1 + x) \) using the series for \( \cos x \) and \( \ln(1 + x) \). Step 3: Apply the limit.
After simplifying the expression and equating the limit to 2, we solve for \( a^2 + b^2 + c^2 \). Final Answer: \[ \boxed{10}. \]