1. Stating the problem: We want to find the limit $$\lim_{x \to 0} \frac{\sqrt{x} - \sin x}{\ln x}$$. 2. Important note: The limit involves $x \to 0$, but $\ln x$ is only defined for $x > 0$. So we consider the limit as $x \to 0^+$. 3. Check behavior of numerator and denominator as $x \to 0^+$: - $\sqrt{x} \to 0$ - $\sin x \to 0$ - $\ln x \to -\infty$ 4. Use series expansions near 0: - $\sqrt{x} = x^{1/2}$ - $\sin x = x - \frac{x^3}{6} + O(x^5)$ - $\ln x = \ln(1 + (x-1))$, but since $x \to 0^+$, $\ln x \to -\infty$ 5. Numerator approximation: $$\sqrt{x} - \sin x = x^{1/2} - \left(x - \frac{x^3}{6} + \cdots \right) = x^{1/2} - x + \frac{x^3}{6} + \cdots$$ 6. As $x \to 0^+$, $x^{1/2}$ dominates $x$ and higher powers, so numerator behaves like $x^{1/2}$. 7. Denominator $\ln x \to -\infty$. 8. So the limit behaves like: $$\lim_{x \to 0^+} \frac{x^{1/2}}{\ln x}$$ 9. Since numerator $\to 0$ and denominator $\to -\infty$, the fraction tends to 0. 10. Therefore, the limit is: $$\boxed{0}$$