Limit Expression E4E4F2

1. **State the problem:** We need to find the limit of the expression $$x - (-1) \cdot \frac{x^3 + 1}{x + 1}$$ as $x$ approaches $-1$. 2. **Rewrite the expression:** The expression can be written as $$x + \frac{x^3 + 1}{x + 1}$$ because $-(-1) = +1$. 3. **Factor the numerator:** Note that $x^3 + 1$ is a sum of cubes, which factors as $$x^3 + 1 = (x + 1)(x^2 - x + 1)$$. 4. **Simplify the fraction:** Substitute the factorization into the fraction: $$\frac{x^3 + 1}{x + 1} = \frac{(x + 1)(x^2 - x + 1)}{x + 1}$$ Cancel the common factor $x + 1$: $$\frac{\cancel{(x + 1)}(x^2 - x + 1)}{\cancel{x + 1}} = x^2 - x + 1$$ 5. **Rewrite the original expression:** Now the expression is $$x + x^2 - x + 1 = x^2 + 1$$ 6. **Evaluate the limit:** Substitute $x = -1$ into $x^2 + 1$: $$(-1)^2 + 1 = 1 + 1 = 2$$ **Final answer:** $$\lim_{x \to -1} \left(x - (-1) \cdot \frac{x^3 + 1}{x + 1}\right) = 2$$