Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{e^{x} - a - \log(1+x)}{\sin x} = 0, { then find } a. \]

• A. \( 2 \)
• B. \( 0 \)
• C. \( -1 \)
• D. \( 1 \)

Hint:

For limits involving exponential, logarithmic, and trigonometric functions, Taylor series expansions are a powerful tool to simplify expressions and find limits.

Answer 1

The Correct Option is D

Step 1: Use the Taylor series expansions of \( e^x \), \( \log(1+x) \), and \( \sin x \) around \( x = 0 \) to simplify the expression: \[ e^x \approx 1 + x + \frac{x^2}{2}, \quad \log(1+x) \approx x - \frac{x^2}{2}, \quad \sin x \approx x - \frac{x^3}{6}. \] Step 2: Substitute these approximations into the limit: \[ \lim_{x \to 0} \frac{(1 + x + \frac{x^2}{2}) - a - (x - \frac{x^2}{2})}{x - \frac{x^3}{6}} = \lim_{x \to 0} \frac{1 - a + x^2}{x - \frac{x^3}{6}}. \] Step 3: For the limit to be zero, the numerator must not have a constant term other than zero when the denominator is approximated by \( x \), hence \( 1 - a = 0 \). 
Step 4: Solving \( 1 - a = 0 \) gives \( a = 1 \).