Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 10

a.) Determine what is wrong in the equation $\displaystyle \frac{x^2+x-6}{x-2} = x+3$

The function in the left side is defined for all values of $x$ except for $x=2$. However, the function on the right side is
on every values of $x$

b.) Prove that the equations $\lim\limits_{x \rightarrow 2} \displaystyle \frac{x^2+x-6}{x-2} = \lim\limits_{x \rightarrow 2} (x+3)$ is correct.


$\begin{equation} \begin{aligned} \lim\limits_{x \rightarrow 2} \displaystyle \frac{x^2+x-6}{x-2} & = \lim\limits_{x \rightarrow 2} \displaystyle \frac{(x+3)\cancel{(x-2)}}{\cancel{x-2}} & & \text{(By Factoring)}\\ \lim\limits_{x \rightarrow 2} \displaystyle \frac{x^2+x-6}{x-2} & = \lim\limits_{x \rightarrow 2} \displaystyle (x+3) \end{aligned} \end{equation}$