Question

\(\lim_{x \to 0} \frac{e - (1 + 2x)^{\frac{1}{2x}}}{x} \quad \text{is equal to:}\)

• A. e
• B. \( -\frac{2}{e} \)
• C. 0
• D. \( e - e^2 \)

Answer 1

The Correct Option is A

The initial limit is given as:

\[ \lim_{x \to 0} \frac{e - e^{\frac{1}{2x} \ln(1 + 2x)}}{x} \]

This expression can be rewritten as:

\[ = \lim_{x \to 0} (-e) \left( \frac{e^{\frac{\ln(1 + 2x)}{2x}} - 1}{x} \right) \]

Applying the standard exponential limit property:

\[ \lim_{x \to 0} \frac{e^{f(x)} - 1}{x} = e^{0} \cdot f'(0) \]

Where \( f(x) = \frac{\ln(1 + 2x)}{2x} \), its derivative is:

\[ f'(x) = \frac{\ln(1 + 2x) - 2x}{2x^2} \]

Therefore, the limit becomes:

\[ \lim_{x \to 0} (-e) \cdot \frac{\ln(1 + 2x) - 2x}{2x^2} \]

Evaluating the limit as \( x \to 0 \):

\[ (-e) \times (-1) \times \frac{4}{2 \times 2} = e \]

The final result is:

\[ \boxed{e} \]