1. **Problem Statement:** Find the limit \( \lim_{x \to -3} \sqrt[3]{\frac{x+3}{x^3+27}} \). 2. **Recall the formula and rules:** - The cube root function is continuous everywhere. - The limit of a quotient is the quotient of the limits if the denominator limit is not zero. - Factor the denominator: \(x^3 + 27 = (x+3)(x^2 - 3x + 9)\). 3. **Evaluate the limit:** \[ \lim_{x \to -3} \sqrt[3]{\frac{x+3}{x^3+27}} = \lim_{x \to -3} \sqrt[3]{\frac{x+3}{(x+3)(x^2 - 3x + 9)}} \] 4. **Simplify the expression inside the cube root:** \[ = \lim_{x \to -3} \sqrt[3]{\frac{1}{x^2 - 3x + 9}} \] 5. **Substitute \(x = -3\):** \[ = \sqrt[3]{\frac{1}{(-3)^2 - 3(-3) + 9}} = \sqrt[3]{\frac{1}{9 + 9 + 9}} = \sqrt[3]{\frac{1}{27}} = \sqrt[3]{\frac{1}{27}} = \frac{1}{3} \] **Final answer:** \( \boxed{\frac{1}{3}} \)