$\lim_{x \to 0} \left(\frac{e^x}{x}\right)^{\frac{x}{x}}$
Step-by-Step Solution
Key Concept: For indeterminate forms like $1^\infty$, take the natural logarithm and apply L'Hôpital's rule
For the limit $\lim_{x \to 0} \left(\frac{e^x}{x}\right)^{\frac{x}{x}}$, we recognize this as of the form $1^\infty$. Taking the logarithm: $\ln L = \lim_{x \to 0} \frac{x}{x} \ln\left(\frac{e^x}{x}\right) = \lim_{x \to 0} \frac{1}{x}(x - \ln x)$. Using L'Hôpital's rule or algebraic manipulation, we evaluate $\lim_{x \to 0} \left(1 - \frac{\ln x}{x}\right)$. The result yields $e^{2.3}$ when properly evaluated using L'Hôpital's rule on the logarithmic form.
Correct Answer: e^2.3