Question

$\underset{x\to -3}{\lim }\frac{x^2-9}{x^2-x-12}$find the limit

Short answer

$\underset{x\to -3}{\lim }\frac{x^2-9}{x^2-x-12}=\frac{6}{7}$

Step-by-step solution

Step 1. Given information 

$\underset{x\to -3}{\lim }\frac{x^2-9}{x^2-x-12}$

Step 2. Domain of the function

We can find limit of rational function by $\underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)$if c is in the domain of f(x) .So to find the domain of f(x)=$\frac{x^2-9}{x^2-x-12}$we have to put denominator =0

role="math" localid="1647084893764" $$\begin{gathered} x^2-x-12 \\ x=-3 \\ x=4 \end{gathered}$$

So domain of the f(x) is all real number except 4and -3.

Step 3. limit of the function

As here c= -3which is not in the domain . So we have to simplify the f(x)

$$\begin{gathered} \underset{x\to -3}{\lim }\frac{x^2-9}{x^2-x-12}=\underset{x\to -3}{\lim }\frac{(x-3)(x+3)}{(x-4)(x+3)} \\ =\underset{x\to -3}{\lim }\frac{x-3}{x-4} \\ =\frac{-3-3}{-3-4}=\frac{6}{7} \end{gathered}$$