Question

question 13 evaluate the limit $lim_{x
ightarrow - 9}\frac{x^{2}+16x + 63}{x + 9}$

Explanation:

Step1: Factor the numerator

We factor \(x^{2}+16x + 63\) as \((x + 7)(x+9)\). So the limit becomes \(\lim_{x
ightarrow - 9}\frac{(x + 7)(x + 9)}{x+9}\).

Step2: Cancel out common factors

Since \(x
eq - 9\) when taking the limit (we are approaching - 9), we can cancel out the \((x + 9)\) terms. The expression simplifies to \(\lim_{x
ightarrow - 9}(x + 7)\).

Step3: Substitute the value of x

Substitute \(x=-9\) into \(x + 7\). We get \(-9+7=-2\).

Answer:

\(-2\)