\( \lim _ { x \rightarrow 0 ^ { + } } \frac { \sqrt { \ln ^ { 2 } x + 3 } + 2 \ln x } { \sqrt { \ln ^ { 2 } x + 1 } + \sqrt { \ln ^ { 2 } x + 2 } } \)

To evaluate the limit \[ L = \lim_{x \rightarrow 0^+} \frac{\sqrt{\ln^2 x + 3} + 2 \ln x}{\sqrt{\ln^2 x + 1} + \sqrt{\ln^2 x + 2}}, \] we start by analyzing the behavior of \(\ln x\) as \(x\) approaches \(0^+\). We know that \(\ln x \to -\infty\) as \(x \to 0^+\). Therefore, we can denote \(y = \ln x\), which implies that as \(x \to 0^+\), \(y \to -\infty\). Substituting \(y\) into the limit, we have: \[ L = \lim_{y \to -\infty} \frac{\sqrt{y^2 + 3} + 2y}{\sqrt{y^2 + 1} + \sqrt{y^2 + 2}}. \] Next, we simplify the expressions in the limit. As \(y \to -\infty\), we can factor \(y^2\) out of the square roots: \[ \sqrt{y^2 + 3} = \sqrt{y^2(1 + \frac{3}{y^2})} = |y|\sqrt{1 + \frac{3}{y^2}} = -y\sqrt{1 + \frac{3}{y^2}}, \] \[ \sqrt{y^2 + 1} = \sqrt{y^2(1 + \frac{1}{y^2})} = -y\sqrt{1 + \frac{1}{y^2}}, \] \[ \sqrt{y^2 + 2} = \sqrt{y^2(1 + \frac{2}{y^2})} = -y\sqrt{1 + \frac{2}{y^2}}. \] Substituting these back into the limit gives: \[ L = \lim_{y \to -\infty} \frac{-y\sqrt{1 + \frac{3}{y^2}} + 2y}{-y\sqrt{1 + \frac{1}{y^2}} - y\sqrt{1 + \frac{2}{y^2}}}. \] We can factor out \(-y\) from both the numerator and the denominator: \[ L = \lim_{y \to -\infty} \frac{-y\left(\sqrt{1 + \frac{3}{y^2}} - 2\right)}{-y\left(\sqrt{1 + \frac{1}{y^2}} + \sqrt{1 + \frac{2}{y^2}}\right)} = \lim_{y \to -\infty} \frac{\sqrt{1 + \frac{3}{y^2}} - 2}{\sqrt{1 + \frac{1}{y^2}} + \sqrt{1 + \frac{2}{y^2}}}. \] As \(y \to -\infty\), \(\frac{3}{y^2} \to 0\), \(\frac{1}{y^2} \to 0\), and \(\frac{2}{y^2} \to 0\). Thus, we have: \[ \sqrt{1 + \frac{3}{y^2}} \to 1, \quad \sqrt{1 + \frac{1}{y^2}} \to 1, \quad \sqrt{1 + \frac{2}{y^2}} \to 1. \] Substituting these limits into our expression gives: \[ L = \frac{1 - 2}{1 + 1} = \frac{-1}{2} = -\frac{1}{2}. \] Thus, the final result is: \[ \boxed{-\frac{1}{2}}. \]