Chapter 2.3, Problem 8E

(a) What is wrong with the following equation?

$\frac{x^2+x-6}{x-2}=x+3$

(b) In view of part (a). explain why the equation

$\underset{x\to 2}{\lim }\frac{x^2+x-6}{x-2}=\underset{x\to 2}{\lim }x+3$

is correct.

To determine

To explain: The equation $\frac{x^2+x-6}{x-2}=x+3$ is incorrect.

Explanation of Solution

Let the function $f\left(x\right)=\frac{x^2+x-6}{x-2}$ and $g\left(x\right)=x+3$.

Construct the table of $f\left(x\right)$ and $g\left(x\right)$ for some arbitrary values of x.

x	$f\left(x\right)$	$g\left(x\right)$
−3	$f\left(-3\right)=\frac{{\left(-3\right)}^2+\left(-3\right)-6}{\left(-3\right)-2}=0$	$g\left(-3\right)=-3+3=0$
−2	$f\left(-2\right)=\frac{{\left(-2\right)}^2+\left(-2\right)-6}{\left(-2\right)-2}=1$	$g\left(-2\right)=-2+3=1$
−1	$f\left(-1\right)=\frac{{\left(-1\right)}^2+\left(-1\right)-6}{\left(-1\right)-2}=2$	$g\left(-1\right)=-1+3=2$
0	$f\left(0\right)=\frac{{\left(0\right)}^2+\left(0\right)-6}{\left(0\right)-2}=3$	$g\left(0\right)=0+3=3$
1	$f\left(1\right)=\frac{{\left(1\right)}^2+\left(1\right)-6}{\left(1\right)-2}=4$	$g\left(1\right)=1+3=4$
2	$f\left(2\right)=\frac{{\left(2\right)}^2+\left(2\right)-6}{\left(2\right)-2}=\frac{0}{0}$	$g\left(2\right)=2+3=5$
3	$f\left(3\right)=\frac{{\left(3\right)}^2+\left(3\right)-6}{\left(3\right)-2}=6$	$g\left(3\right)=3+3=6$

From the table, it is observed that, $f\left(x\right)$ is not defined at $x=2$ but $g\left(x\right)$  is defined at $x=2$.

The domain of $f\left(x\right)=\left\{\frac{x^2+x-6}{x-2}:x\in ℝ\text{ except }x=2\right\}$ and the domain of $g\left(x\right)=\left\{x+3:x\in ℝ\right\}$.

In general, the equation holds for all x not equal to 2. That is, $f\left(x\right)=g\left(x\right)$ for $x\in ℝ$ except $x=2$.

Therefore, $f\left(x\right)$ and $g\left(x\right)$ are not equal.

That is, $\frac{x^2+x-6}{x-2}\ne x+3$.

To determine

To explain: The equation $\underset{x\to 2}{\lim }\frac{x^2+x-6}{x-2}=\underset{x\to 2}{\lim }\left(x+3\right)$ is correct.

Explanation of Solution

From part (a), the equation holds for all real numbers except 2.

That is, $\frac{x^2+x-6}{x-2}=x+3$ is true when $x\ne 2$.

$\begin{array}{c}\underset{x\to 2}{\lim }\frac{x^2+x-6}{x-2}=\underset{x\to 2}{\lim }\frac{\left(x-2\right)\left(x+3\right)}{x-2}\\ =\underset{x\to 2}{\lim }\left(x+3\right)\end{array}$

Thus, it can be concluded that both the limit functions are equal.

That is, $\underset{x\to 2}{\lim }\frac{x^2+x-6}{x-2}=\underset{x\to 2}{\lim }\left(x+3\right)$.