Answer
(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 $x=2$.
Work Step by Step
(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.
