Question

\( \lim_{x \to 0} \frac{\log(1 + 3x^2){x(e^{5x} - 1)} = \)}

• A. \( \frac{3}{5} \)
• B. \( \frac{5}{3} \)
• C. \( \frac{-3}{5} \)
• D. \( \frac{-5}{3} \)
• E. \( 1 \)

Hint:

For limits involving \( \log(1+kx^n) \) and \( (e^{mx}-1) \), you can replace the terms with their first-order approximations \( kx^n \) and \( mx \) respectively for quick evaluation. \( \frac{3x^2}{x(5x)} = \frac{3}{5} \).

Answer 1

The Correct Option is A

Concept: We use the standard logarithmic and exponential limits: \[ \lim_{u \to 0} \frac{\log(1 + u)}{u} = 1 \quad \text{and} \quad \lim_{v \to 0} \frac{e^v - 1}{v} = 1 \] We will manipulate the given expression to match these forms.

Step 1: Restructure the expression.
Multiply and divide by terms to create the standard forms: \[ \lim_{x \to 0} \left[ \frac{\log(1 + 3x^2)}{3x^2} \cdot \frac{3x^2}{x(e^{5x} - 1)} \right] \] \[ \lim_{x \to 0} \left[ \frac{\log(1 + 3x^2)}{3x^2} \cdot \frac{3x}{e^{5x} - 1} \right] \]

Step 2: Apply standard limits.
As \( x \to 0 \), the term \( \frac{\log(1 + 3x^2)}{3x^2} \to 1 \). For the second part, manipulate it to match the \( e^v \) limit: \[ \frac{3x}{e^{5x} - 1} = \frac{3}{5} \cdot \frac{5x}{e^{5x} - 1} \] As \( 5x \to 0 \), the term \( \frac{5x}{e^{5x} - 1} \to 1 \).

Step 3: Final Calculation.
\[ \text{Limit} = 1 \cdot \frac{3}{5} \cdot 1 = \frac{3}{5} \]