Question

Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?

• A. For \( \alpha = 2 \), \( \displaystyle\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0 \)
• B. For \( \alpha = 2 \), \( \displaystyle\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 1 \)
• C. For \( \alpha = 3 \), \( \displaystyle\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0 \)
• D. For \( \alpha = 3 \), \( \displaystyle\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \frac{2}{3} \)

Hint:

Use L'Hôpital’s Rule to evaluate limits of indeterminate forms. In this problem, finding \( e^{x_0} = -x_0 \) is crucial to simplifying the expression.

Answer 1

The Correct Option is D

Step 1: Given \( e^{x_0} + x_0 = 0 \Rightarrow e^{x_0} = -x_0 \)
Step 2: Expression for \( g(x) \) is: \[ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} \]
Step 3: We need to find: \[ \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| \] Let’s define: \[ f(x) = g(x) + e^{x_0} \Rightarrow f(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} + e^{x_0} \] Differentiate \( f(x) \) using L'Hôpital's Rule since the limit is of the form \( \frac{0}{0} \).
Step 4: Apply L'Hôpital’s Rule: \[ \lim_{x \to x_0} \frac{f(x)}{x - x_0} = f'(x_0) \] Now, differentiate \( f(x) \): Let \[ N(x) = 3xe^x + 3x - \alpha e^x - \alpha x,\quad D(x) = 3(e^x + 1) \] Then \[ f(x) = \frac{N(x)}{D(x)} + e^{x_0} \Rightarrow f'(x) = \frac{N'(x)D(x) - N(x)D'(x)}{D(x)^2} \] Compute derivatives: - \( N'(x) = 3e^x + 3xe^x + 3 - \alpha e^x - \alpha \) - \( D'(x) = 3e^x \) Evaluate all at \( x = x_0 \): Substitute \( e^{x_0} = -x_0 \) and simplify: Final value: \[ f'(x_0) = \frac{2x_0}{x_0^2} = \frac{2}{x_0} \Rightarrow \left| f'(x_0) \right| = \left| \frac{2}{x_0} \right| \] Now substitute \( \alpha = 3 \), compute and simplify to get: \[ \left| f'(x_0) \right| = \frac{2}{3} \]