$\lim_{x \to \infty} \sqrt[3]{(x+a)(x+b)(x+c)} - x =$
Step-by-Step Solution
Key Concept: Use the algebraic identity for difference of cubes to reduce the limit to a rational form.
For $\lim_{x \to c} [(x+a)(x+b)(x+c)]^{1/3} - x$, we use the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$. Writing the limit with $a = [(x+a)(x+b)(x+c)]^{1/3}$ and $b = x$, the numerator becomes $[x^3 + (a+b+c)x^2 + (ab+bc+ca)x + abc] - x^3$. The limit evaluates to $\frac{a+b+c}{3}$ using the approximation for small perturbations.
Correct Answer: 2