Question

question 3 · 1 point evaluate lim(x→0) ((-3x - 6)^2 - 36)/(-9x). submit an exact answer. provide your answer below:

Explanation:

Step1: Expand the numerator

Expand $(-3x - 6)^2$ using $(a + b)^2=a^{2}+2ab + b^{2}$, where $a=-3x$ and $b = - 6$. So $(-3x - 6)^2=(-3x)^2+2(-3x)(-6)+(-6)^2=9x^{2}+36x + 36$. Then the numerator $(-3x - 6)^2-36=9x^{2}+36x+36 - 36=9x^{2}+36x$.

Step2: Simplify the fraction

The original limit $\lim_{x
ightarrow0}\frac{(-3x - 6)^2-36}{-9x}=\lim_{x
ightarrow0}\frac{9x^{2}+36x}{-9x}$. Factor out $x$ from the numerator: $\lim_{x
ightarrow0}\frac{x(9x + 36)}{-9x}$. Cancel out the non - zero factor $x$ (since $x
ightarrow0$ but $x
eq0$ during the limit process), we get $\lim_{x
ightarrow0}\frac{9x + 36}{-9}$.

Step3: Evaluate the limit

Substitute $x = 0$ into $\frac{9x+36}{-9}$. We have $\frac{9\times0+36}{-9}=\frac{36}{-9}=-4$.

Answer:

$-4$