1. **Stating the problem:** We want to find the limit $$\lim_{x \to 0^+} \log \sqrt{2x}$$ where $x > 0$. 2. **Recall the logarithm and root properties:** - The square root can be written as a power: $$\sqrt{2x} = (2x)^{\frac{1}{2}}$$. - The logarithm of a power is the exponent times the logarithm: $$\log (a^b) = b \log a$$. 3. **Rewrite the expression using these properties:** $$\log \sqrt{2x} = \log (2x)^{\frac{1}{2}} = \frac{1}{2} \log (2x)$$. 4. **Use the logarithm product rule:** $$\log (2x) = \log 2 + \log x$$. 5. **Substitute back:** $$\frac{1}{2} \log (2x) = \frac{1}{2} (\log 2 + \log x) = \frac{1}{2} \log 2 + \frac{1}{2} \log x$$. 6. **Evaluate the limit as $x \to 0^+$:** - We know $$\lim_{x \to 0^+} \log x = -\infty$$ because the logarithm approaches negative infinity as its argument approaches zero from the right. - The term $$\frac{1}{2} \log 2$$ is constant. 7. **Therefore:** $$\lim_{x \to 0^+} \log \sqrt{2x} = \frac{1}{2} \log 2 + \frac{1}{2} \lim_{x \to 0^+} \log x = \frac{1}{2} \log 2 + \frac{1}{2} (-\infty) = -\infty$$. **Final answer:** $$\boxed{-\infty}$$