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

Answer: (A)

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: $e.$