Question

find the limit. (limlimits_{x \to -9} \frac{x^2 + 7x - 18}{x + 9})

Explanation:

Step1: Factor the numerator

We factor the quadratic expression \(x^2 + 7x - 18\). We need two numbers that multiply to \(-18\) and add up to \(7\). The numbers are \(9\) and \(-2\). So, \(x^2 + 7x - 18=(x + 9)(x - 2)\).
The limit becomes \(\lim_{x
ightarrow - 9}\frac{(x + 9)(x - 2)}{x + 9}\)

Step2: Cancel the common factor

Since \(x
ightarrow - 9\) but \(x
eq - 9\) (we are taking the limit, not evaluating at \(x=-9\)), we can cancel the common factor \(x + 9\) from the numerator and the denominator.
After canceling, we get \(\lim_{x
ightarrow - 9}(x - 2)\)

Step3: Evaluate the limit

Now we substitute \(x=-9\) into the expression \(x - 2\).
\(-9-2=-11\)

Answer:

\(-11\)