1. Stating the problem: Evaluate the limit $$\lim_{x \to 0} \frac{\sqrt[3]{1 + cx - 1}}{x} = \lim_{x \to 0} \frac{\sqrt[3]{cx}}{x}.$$ 2. Simplify the expression inside the cube root: Since $$1 + cx - 1 = cx,$$ the limit becomes $$\lim_{x \to 0} \frac{\sqrt[3]{cx}}{x}.$$ 3. Rewrite the cube root as a power: $$\sqrt[3]{cx} = (cx)^{1/3} = c^{1/3} x^{1/3}.$$ 4. Substitute back into the limit: $$\lim_{x \to 0} \frac{c^{1/3} x^{1/3}}{x} = c^{1/3} \lim_{x \to 0} \frac{x^{1/3}}{x} = c^{1/3} \lim_{x \to 0} x^{1/3 - 1} = c^{1/3} \lim_{x \to 0} x^{-2/3}.$$ 5. Analyze the limit: As $$x \to 0,$$ $$x^{-2/3} = \frac{1}{x^{2/3}}$$ approaches infinity. Therefore, the limit does not exist (it diverges to infinity or negative infinity depending on the direction and sign of $$c$$). 6. Conclusion: The limit $$\lim_{x \to 0} \frac{\sqrt[3]{1+cx-1}}{x}$$ diverges (does not exist) for any nonzero constant $$c$$. If $$c = 0,$$ the limit trivially equals 0.