1. **State the problem:** We need to find the limit $$\lim_{x \to -1} \frac{x^3 + 3x^2 + x - 1}{x + 1}$$ 2. **Check direct substitution:** Substitute $x = -1$ into numerator and denominator: Numerator: $(-1)^3 + 3(-1)^2 + (-1) - 1 = -1 + 3 - 1 - 1 = 0$ Denominator: $-1 + 1 = 0$ Since direct substitution gives $\frac{0}{0}$, an indeterminate form, we need to simplify. 3. **Factor the numerator:** Use polynomial division or factor by synthetic division to divide $x^3 + 3x^2 + x - 1$ by $x + 1$. Divide: $$x^3 + 3x^2 + x - 1 \div (x + 1)$$ - First term: $x^2$ because $x^2 \times (x + 1) = x^3 + x^2$ - Subtract: $(x^3 + 3x^2) - (x^3 + x^2) = 2x^2$ - Bring down $+ x$: now $2x^2 + x$ - Next term: $+ 2x$ because $2x \times (x + 1) = 2x^2 + 2x$ - Subtract: $(2x^2 + x) - (2x^2 + 2x) = -x$ - Bring down $-1$: now $-x - 1$ - Next term: $-1$ because $-1 \times (x + 1) = -x - 1$ - Subtract: $(-x - 1) - (-x - 1) = 0$ So, $$x^3 + 3x^2 + x - 1 = (x + 1)(x^2 + 2x - 1)$$ 4. **Simplify the limit expression:** $$\frac{x^3 + 3x^2 + x - 1}{x + 1} = \frac{(x + 1)(x^2 + 2x - 1)}{x + 1} = x^2 + 2x - 1, \quad x \neq -1$$ 5. **Evaluate the limit:** Now substitute $x = -1$ into the simplified expression: $$(-1)^2 + 2(-1) - 1 = 1 - 2 - 1 = -2$$ **Final answer:** $$\lim_{x \to -1} \frac{x^3 + 3x^2 + x - 1}{x + 1} = -2$$