Answer is: option2

Solution:

Let:

\[ y = (1 + x)^{\frac{1}{x}} \]

Taking the natural logarithm on both sides:

\[ \ln y = \frac{1}{x} \ln (1 + x) \]

Now, we need to evaluate:

\[ \lim_{x\to0} \frac{\ln(1 + x)}{x} \]

Since both the numerator and denominator approach 0 as \( x \to 0 \), we apply L'Hôpital's Rule, differentiating the numerator and denominator separately:

\[ \lim_{x\to0} \frac{\ln(1 + x)}{x} = \lim_{x\to0} \frac{\frac{d}{dx} \ln(1 + x)}{\frac{d}{dx} x} \]

Since the derivative of \( \ln(1 + x) \) is \( \frac{1}{1+x} \) and the derivative of \( x \) is 1, we get:

\[ \lim_{x\to0} \frac{1}{1 + x} = 1 \]

Since \( \lim_{x\to0} \ln y = 1 \), we exponentiate both sides:

\[ \lim_{x\to0} y = e^1 = e \]

The correct answer is \( e \).