## Calculus: Early Transcendentals 8th Edition

(a) The equation has one side that is not defined with $x=2$, while the other side is still defined with $x=2$. (b) The equation is correct because when computing the limit as $x$ approaches $2$, we never actually consider the situation when $x=2$.
(a) Consider the equation. $\frac{x^2+x-6}{x-2}=x+3$ We see that - In the left side of the equation, $x\ne2$ or else, the left side would not be defined. - In the right side of the equation, $x+3$ is defined with $x=2$. Therefore, the equation has one side that is not defined with $x=2$, while the other side is still defined with $x=2$. (b) We see that $\lim\limits_{x\to 2}\frac{x^2+x-6}{x-2}$ $=\lim\limits_{x\to 2}\frac{(x-2)(x+3)}{x-2}$ $=\lim\limits_{x\to 2} (x+3)$ While the left-hand side is undefined for $x=2$, when we compute the limit as $x$ approaches $2$, we never consider the situation when $x=2$, but only the value of $x$ near $2$. Also, both sides are defined any $x\ne2$. Therefore, the equation is still correct.