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 natural logarithm function $\ln x$ is only defined for $x > 0$, and as $x \to 0^+$, $\ln x \to -\infty$. 3. Analyze the numerator as $x \to 0^+$: - $\sqrt{x} \to 0$ - $\sin x \to 0$ So numerator $\to 0$. 4. Analyze the denominator as $x \to 0^+$: - $\ln x \to -\infty$ 5. So the expression behaves like $\frac{0}{-\infty}$ which tends to 0. 6. Therefore, the limit is: $$\lim_{x \to 0^+} \frac{\sqrt{x} + \sin x}{\ln x} = 0$$. Note: The limit from the left side $x \to 0^-$ is not defined because $\ln x$ is undefined for $x \leq 0$.