Find the limit. \[ \lim (1+2 x)^{\frac{1}{2 \ln x}} \]

To find the limit \[ L = \lim_{x \to 0^+} (1 + 2x)^{\frac{1}{2 \ln x}}, \] we start by rewriting the expression in a more manageable form. We can take the natural logarithm of \(L\): \[ \ln L = \lim_{x \to 0^+} \frac{1}{2 \ln x} \ln(1 + 2x). \] Next, we need to analyze the behavior of \(\ln(1 + 2x)\) as \(x\) approaches \(0^+\). Using the Taylor expansion for \(\ln(1 + u)\) around \(u = 0\), we have: \[ \ln(1 + 2x) \approx 2x \quad \text{as } x \to 0^+. \] Thus, we can approximate: \[ \ln(1 + 2x) \sim 2x. \] Substituting this approximation into our limit gives: \[ \ln L = \lim_{x \to 0^+} \frac{1}{2 \ln x} \cdot 2x = \lim_{x \to 0^+} \frac{x}{\ln x}. \] Now, we need to evaluate the limit \(\lim_{x \to 0^+} \frac{x}{\ln x}\). As \(x \to 0^+\), \(\ln x \to -\infty\), and thus the fraction \(\frac{x}{\ln x}\) approaches \(0\) because \(x\) approaches \(0\) much faster than \(\ln x\) approaches \(-\infty\). Therefore, we have: \[ \lim_{x \to 0^+} \frac{x}{\ln x} = 0. \] This implies: \[ \ln L = 0. \] Exponentiating both sides gives: \[ L = e^0 = 1. \] Thus, the limit is \[ \boxed{1}. \]