1. **State the problem:** Find the limit $$\lim_{x \to 0^-} \frac{\sqrt{1+2x}-1}{3x}$$. 2. **Recall the formula and approach:** When direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, we use algebraic manipulation such as rationalizing the numerator. 3. **Check direct substitution:** Substitute $x=0$: $$\frac{\sqrt{1+2\cdot0}-1}{3\cdot0} = \frac{\sqrt{1}-1}{0} = \frac{0}{0}$$ which is indeterminate. 4. **Rationalize the numerator:** Multiply numerator and denominator by the conjugate $$\sqrt{1+2x}+1$$: $$\frac{\sqrt{1+2x}-1}{3x} \cdot \frac{\sqrt{1+2x}+1}{\sqrt{1+2x}+1} = \frac{(\sqrt{1+2x})^2 - 1^2}{3x(\sqrt{1+2x}+1)} = \frac{1+2x - 1}{3x(\sqrt{1+2x}+1)} = \frac{2x}{3x(\sqrt{1+2x}+1)}$$ 5. **Cancel common factor $x$:** $$\frac{2\cancel{x}}{3\cancel{x}(\sqrt{1+2x}+1)} = \frac{2}{3(\sqrt{1+2x}+1)}$$ 6. **Evaluate the limit by substituting $x=0$ now:** $$\frac{2}{3(\sqrt{1+0}+1)} = \frac{2}{3(1+1)} = \frac{2}{6} = \frac{1}{3}$$ **Final answer:** $$\lim_{x \to 0^-} \frac{\sqrt{1+2x}-1}{3x} = \frac{1}{3}$$