Answer is: option1
\( e \)Solution:
Define:
\[ L = \lim_{x \to \infty} \left( e^x + x \right)^{\frac{1}{x}} \]
Taking the natural logarithm:
\[ \ln L = \lim_{x \to \infty} \frac{1}{x} \ln (e^x + x) \]
For large \( x \), the term \( e^x \) dominates over \( x \), so we approximate:
\[ e^x + x \approx e^x \]
Taking the logarithm:
\[ \ln(e^x + x) \approx \ln e^x = x \]
Thus:
\[ \ln L = \lim_{x \to \infty} \frac{1}{x} \cdot x = 1 \]
Since \( \ln L = 1 \), we exponentiate both sides:
\[ L = e^1 = e \]
Correct Answer: \( e \)