Question

In Problems 7– 42, find each limit algebraically.

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

Short answer

The answer is$\frac{5}{4}.$

Step-by-step solution

Step. 1 Given Information

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

$f\left(x\right)=\frac{x^2+x-6}{x^2+2x-3}$

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

$x^2+x-6={(-3)}^2+(-3)-6=9-3-6=9-9=0.$

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

$x^2+2x-3={(-3)}^2+2(-3)-3=9-6-3=9-9=0.$

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

Step. 2 Factorizing

Numerator, $x^2+x-6=(x+3)(x-2)$.

Denominator, $x^2+2x-3=(x+3)(x-1)$.

So,

$$\begin{gathered} \underset{x\to -3}{\lim }\frac{x^2+x-6}{x^2+2x-3}=\underset{x\to -3}{\lim }\frac{(x+3)(x-2)}{(x+3)(x-1)} \\ =\underset{x\to -3}{\lim }\frac{x-2}{x-1}. \end{gathered}$$

Now we can put the limit directly.

Step. 3 Final calculation of the limit 

$\underset{x\to -3}{\lim }\frac{x-2}{x-1}=\frac{-3-2}{-3-1}=\frac{-5}{-4}=\frac{5}{4}.$