$$\\lim_{x\\to -7}\\frac{-3x - 12}{\\sqrt{x + 8}-6}$$
$$\\lim_{x\\to -7}\\frac{-3x - 12}{\\sqrt{x + 8}-6}$$
Answer
Explanation:
Step1: Rationalize the denominator
Multiply the numerator and denominator by $\sqrt{x + 8}+6$. $$\lim_{x\rightarrow - 7}\frac{(-3x - 12)(\sqrt{x + 8}+6)}{(\sqrt{x + 8}-6)(\sqrt{x + 8}+6)}$$ Since $(a - b)(a + b)=a^{2}-b^{2}$, the denominator becomes $(x + 8)-36=x - 28$. So we have $\lim_{x\rightarrow - 7}\frac{(-3x - 12)(\sqrt{x + 8}+6)}{x - 28}$. Factor the numerator: $-3x-12=-3(x + 4)$. The limit is now $\lim_{x\rightarrow - 7}\frac{-3(x + 4)(\sqrt{x + 8}+6)}{x - 28}$.
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})