1. **State the problem:** We are given the equation $$\frac{x^2 + x - 6}{x - 2} = x + 3$$ and asked what is wrong with it. 2. **Analyze the equation:** The numerator can be factored: $$x^2 + x - 6 = (x + 3)(x - 2)$$ So the left side becomes: $$\frac{(x + 3)(x - 2)}{x - 2}$$ 3. **Simplify the expression:** For all $x \neq 2$, we can cancel the common factor $x - 2$: $$\frac{\cancel{(x - 2)}(x + 3)}{\cancel{x - 2}} = x + 3$$ 4. **What is wrong?** The original equation is not true for $x = 2$ because the denominator is zero, making the expression undefined. The simplification assumes $x \neq 2$. 5. **Explain the limit equation:** The limit $$\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} = \lim_{x \to 2} (x + 3)$$ is correct because the two expressions are equal for all $x$ near 2 except at $x=2$ itself. 6. **Evaluate the limit:** $$\lim_{x \to 2} (x + 3) = 2 + 3 = 5$$ **Final answer:** - The equation $$\frac{x^2 + x - 6}{x - 2} = x + 3$$ is not valid at $x=2$ because the left side is undefined there. - The limit equation is correct because limits consider values arbitrarily close to 2, not at 2 itself.