Question

In Problems 7– 42, find each limit algebraically.

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

Short answer

The answer is$\frac{7}{6}.$

Step-by-step solution

Step. 1 Given Information

Firstly, we check whether the given function is in indeterminant form or not.

$f\left(x\right)=\frac{x^2-x-12}{x^2-9}$

Put $x=-3$in numerator we get,

$x^2-x-12={(-3)}^2-(-3)-12=9+3-12=0.$

Now, put $x=-3$in denominator we get,

$x^2-9={(-3)}^2-9=9-9=0$.

Since both numerator and denominator gives 0 means they both have$x=-3$as their common root.

Step. 2 Factorizing 

Numerator, $x^2-x-12=(x+3)(x-4)$,

Denominator, $x^2-9=(x+3)(x-3)$,

So,

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

Now we can put the limit directly.

Step. 3 Final calculation of the limit

$\underset{x\to -3}{\lim }\frac{x-4}{x-3}=\frac{-3-4}{-3-3}=\frac{-7}{-6}=\frac{7}{6}.$