## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to3}\frac{x^2-9}{x-3}=\lim\limits_{x\to3}(x+3)$$ We consider $$\lim\limits_{x\to3}\frac{x^2-9}{x-3}$$$$=\lim\limits_{x\to3}\frac{(x-3)(x+3)}{x-3}$$ [for $a^2-b^2=(a-b)(a+b)$]$$=\lim\limits_{x\to3}(x+3)$$ Therefore, the statement is true. The reason that this statement is true, while the last one is false, is that when we consider the limit as $x\to3$, we actually never consider the case when $x=3$, but only the value of $x$ surrounding 3. Since, both sides are satisfied for all $x\in R$ except $x=3$, the equation is correct in this time.