Question

suppose that \\(\lim\limits_{x \to -4} p(x) = 8\\), \\(\lim\limits_{x \to -4} r(x) = 0\\), and \\(\lim\limits_{x \to -4} s(x) = -9\\). find the limits in parts (a) through (c) below.

a. \\(\lim\limits_{x \to -4} (p(x) + r(x) + s(x)) = -1\\) (simplify your answer.)

b. \\(\lim\limits_{x \to -4} (p(x) \cdot r(x) \cdot s(x)) = 0\\) (simplify your answer.)

c. \\(\lim\limits_{x \to -4} \frac{-4p(x) + 3r(x)}{s(x)} = -\frac{32}{9}\\) (simplify your answer.)

Explanation:

Part (a)

Step1: Use Sum Rule of Limits

The limit of a sum is the sum of the limits. So, $\lim_{x\to -4}(p(x) + r(x) + s(x))=\lim_{x\to -4}p(x)+\lim_{x\to -4}r(x)+\lim_{x\to -4}s(x)$.

Step2: Substitute Given Limits

We know $\lim_{x\to -4}p(x) = 8$, $\lim_{x\to -4}r(x)=0$, and $\lim_{x\to -4}s(x)= - 9$. Substituting these values: $8 + 0+(-9)=8 - 9=-1$.

Step1: Use Product Rule of Limits

The limit of a product is the product of the limits. So, $\lim_{x\to -4}(p(x)\cdot r(x)\cdot s(x))=\lim_{x\to -4}p(x)\cdot\lim_{x\to -4}r(x)\cdot\lim_{x\to -4}s(x)$.

Step2: Substitute Given Limits

Substitute $\lim_{x\to -4}p(x) = 8$, $\lim_{x\to -4}r(x)=0$, and $\lim_{x\to -4}s(x)= - 9$: $8\times0\times(-9)=0$.

Step1: Use Quotient and Sum Rules

The limit of a quotient is the quotient of the limits (provided the limit of the denominator is non - zero), and the limit of a sum is the sum of the limits. So, $\lim_{x\to -4}\frac{-4p(x)+3r(x)}{s(x)}=\frac{\lim_{x\to -4}(-4p(x)+3r(x))}{\lim_{x\to -4}s(x)}=\frac{- 4\lim_{x\to -4}p(x)+3\lim_{x\to -4}r(x)}{\lim_{x\to -4}s(x)}$.

Step2: Substitute Given Limits

Substitute $\lim_{x\to -4}p(x) = 8$, $\lim_{x\to -4}r(x)=0$, and $\lim_{x\to -4}s(x)= - 9$: $\frac{-4\times8 + 3\times0}{-9}=\frac{-32+0}{-9}=\frac{-32}{-9}=\frac{32}{9}$? Wait, there is a mistake in the original boxed answer. Wait, let's recalculate:

Wait, $\lim_{x\to -4}s(x)=-9$, $\lim_{x\to -4}p(x) = 8$, $\lim_{x\to -4}r(x)=0$.

So, numerator: $-4\times8+3\times0=-32 + 0=-32$

Denominator: $\lim_{x\to -4}s(x)=-9$

So, $\frac{-32}{-9}=\frac{32}{9}$. But the original boxed answer was $-\frac{32}{9}$, which is wrong. Let's do it correctly.

$\lim_{x\to -4}\frac{-4p(x)+3r(x)}{s(x)}=\frac{-4\lim_{x\to -4}p(x)+3\lim_{x\to -4}r(x)}{\lim_{x\to -4}s(x)}=\frac{-4\times8+3\times0}{-9}=\frac{-32}{-9}=\frac{32}{9}$

Wait, maybe the original problem had a typo, but according to the given values:

Answer:

-1

Part (b)