Question

$$\lim_{x\to -7}\frac{-3x - 12}{\sqrt{x + 8}-6}$$

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by $\sqrt{x + 8}+6$.

$$\lim_{x ightarrow - 7}\frac{(-3x - 12)(\sqrt{x + 8}+6)}{(\sqrt{x + 8}-6)(\sqrt{x + 8}+6)}$$

Since $(a - b)(a + b)=a^{2}-b^{2}$, the denominator becomes $(x + 8)-36=x - 28$.
So we have $\lim_{x
ightarrow - 7}\frac{(-3x - 12)(\sqrt{x + 8}+6)}{x - 28}$.
Factor the numerator: $-3x-12=-3(x + 4)$.
The limit is now $\lim_{x
ightarrow - 7}\frac{-3(x + 4)(\sqrt{x + 8}+6)}{x - 28}$.

Step2: Substitute \(x=-7\)

Substitute \(x = - 7\) into the expression:
For the numerator: \(-3(-7 + 4)(\sqrt{-7+8}+6)=-3\times(-3)(1 + 6)=9\times7 = 63\).
For the denominator: \(-7-28=-35\).

Answer:

\(\frac{63}{-35}=-\frac{9}{5}\)