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

We start with the given limit:

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

Rewrite the expression as:

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

Now, apply the standard exponential limit property:

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

Here, \( f(x) = \frac{\ln(1 + 2x)}{2x} \), so

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

Thus,

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

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

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

Hence, the final answer is:

\[ \boxed{e} \]

Answer 2

Let \( L = \lim_{x \to 0} \frac{e - (1 + 2x)^{\frac{1}{2x}}}{x} \).

Rewrite \( (1 + 2x)^{\frac{1}{2x}} \) using logarithms:

\[ (1 + 2x)^{\frac{1}{2x}} = e^{\frac{\ln(1 + 2x)}{2x}}. \]

For small \( x \), use the approximation \( \ln(1 + 2x) \approx 2x \). Thus:

\[ \frac{\ln(1 + 2x)}{2x} \approx 1, \] so \( (1 + 2x)^{\frac{1}{2x}} \approx e^1 = e \).

Expand \( \ln(1 + 2x) \) further using Taylor series:

\[ \ln(1 + 2x) = 2x - 2x^2 + O(x^3), \] so \[ \frac{\ln(1 + 2x)}{2x} = 1 - x + O(x^2). \]

Hence,

\[ (1 + 2x)^{\frac{1}{2x}} = e^{1 - x + O(x^2)} = e \cdot e^{-x} \cdot e^{O(x^2)} \approx e(1 - x + O(x^2)). \]

Subtract from \( e \):

\[ e - (1 + 2x)^{\frac{1}{2x}} \approx e - e(1 - x) = e \cdot x. \]

Divide by \( x \):

\[ \frac{e - (1 + 2x)^{\frac{1}{2x}}}{x} \approx \frac{e \cdot x}{x} = e. \]

Therefore: \[ e. \]