Calculus 1: L'Hôpital's Rule with Indeterminate Forms (0/0, ∞/∞, and 0⁰)

Exact Midterm Problem: Indeterminate Power $0^0$

Evaluate $\lim_{x\to 0^+} x^x$.

Step-by-Step Solution

1. Identify Indeterminate Form: Direct substitution gives $0^0$.
2. Take Natural Logarithm:

$$ \ln L = \lim_{x\to 0^+} \ln(x^x) = \lim_{x\to 0^+} x\ln x = \lim_{x\to 0^+} \frac{\ln x}{1/x} \quad \left(\text{Form } \frac{-\infty}{\infty}\right) $$

1. Apply L'Hôpital's Rule:

$$ \ln L = \lim_{x\to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x\to 0^+} (-x) = 0 $$

1. Exponentiate to Solve for $L$:

$$ \ln L = 0 \implies L = e^0 = \mathbf{1} $$