1. The problem is to analyze the equation $$\frac{x^2 + x - 6}{x - 2} = x + 3$$ and identify what is wrong with it. 2. First, factor the numerator: $$x^2 + x - 6 = (x + 3)(x - 2)$$ 3. Substitute into the fraction: $$\frac{(x + 3)(x - 2)}{x - 2}$$ 4. For all values of $x$ except $x = 2$ (which makes the denominator zero), the $x-2$ terms cancel: $$x + 3$$ 5. So the equation holds for all $x \neq 2$, but it is not defined at $x = 2$ because division by zero is undefined. 6. Therefore, the given equation is wrong if interpreted as holding for $x=2$, but correct for other $x$ values. 7. Part (b) asks why the limit equations $$\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} = \lim_{x \to 2} (x + 3)$$ are correct. 8. When we take limits, we consider values of $x$ close to 2 but not equal to 2. 9. Since the expression simplifies to $x + 3$ for all $x \neq 2$, the limits are equal: $$\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} = \lim_{x \to 2} (x + 3) = 2 + 3 = 5$$ 10. The limit is defined even though the function is undefined at $x=2$ itself, explaining why the limit equality holds. Final answer: - Equation is invalid at $x=2$ as denominator is zero - Limits are equal because expressions are equal everywhere except at $x=2$ where limit cares about approaching values.