Answer :
Let's solve these limit problems step-by-step:
Given:
[tex]\[ \lim_{x \to 3} f(x) = 4 \quad \text{and} \quad \lim_{x \to 3} g(x) = -7 \][/tex]
### Part (a)
[tex]\[ \lim_{x \to 3} [f(x) g(x)] \][/tex]
Using the limit property that states [tex]\(\lim_{x \to a} [f(x) \cdot g(x)] = \left(\lim_{x \to a} f(x)\right) \cdot \left(\lim_{x \to a} g(x)\right)\)[/tex], we have:
[tex]\[ \lim_{x \to 3} [f(x) g(x)] = \left(\lim_{x \to 3} f(x)\right) \cdot \left(\lim_{x \to 3} g(x)\right) = 4 \cdot (-7) = -28 \][/tex]
So,
[tex]\[ \lim_{x \to 3} [f(x) g(x)] = -28 \][/tex]
### Part (b)
[tex]\[ \lim_{x \to 3} [3 f(x) g(x)] \][/tex]
We can factor out the constant 3 from the limit:
[tex]\[ \lim_{x \to 3} [3 f(x) g(x)] = 3 \cdot \lim_{x \to 3} [f(x) g(x)] \][/tex]
From part (a), we know:
[tex]\[ \lim_{x \to 3} [f(x) g(x)] = -28 \][/tex]
So,
[tex]\[ \lim_{x \to 3} [3 f(x) g(x)] = 3 \cdot (-28) = -84 \][/tex]
### Part (c)
[tex]\[ \lim_{x \to 3} [f(x) + 8 g(x)] \][/tex]
Using the limit property that states [tex]\(\lim_{x \to a} [f(x) + g(x)] = \left(\lim_{x \to a} f(x)\right) + \left(\lim_{x \to a} g(x)\right)\)[/tex], we have:
[tex]\[ \lim_{x \to 3} [f(x) + 8 g(x)] = \lim_{x \to 3} f(x) + 8 \cdot \lim_{x \to 3} g(x) \][/tex]
Given,
[tex]\[ \lim_{x \to 3} f(x) = 4 \quad \text{and} \quad \lim_{x \to 3} g(x) = -7 \][/tex]
So,
[tex]\[ \lim_{x \to 3} [f(x) + 8 g(x)] = 4 + 8 \cdot (-7) = 4 + (-56) = 4 - 56 = -52 \][/tex]
### Part (d)
[tex]\[ \lim_{x \to 3} \left[\frac{f(x)}{f(x) - g(x)}\right] \][/tex]
Using the limit properties and given:
[tex]\[ \lim_{x \to 3} f(x) = 4 \quad \text{and} \quad \lim_{x \to 3} g(x) = -7 \][/tex]
First, we find the limit of the denominator:
[tex]\[ \lim_{x \to 3} [f(x) - g(x)] = \lim_{x \to 3} f(x) - \lim_{x \to 3} g(x) = 4 - (-7) = 4 + 7 = 11 \][/tex]
Since the denominator does not approach zero, we can use the limit properties for division:
[tex]\[ \lim_{x \to 3} \left[\frac{f(x)}{f(x) - g(x)}\right] = \frac{\lim_{x \to 3} f(x)}{\lim_{x \to 3} [f(x) - g(x)]} = \frac{4}{11} \][/tex]
So, we have:
[tex]\[ \lim_{x \to 3} \left[\frac{f(x)}{f(x) - g(x)}\right] = \frac{4}{11} \approx 0.36363636363636365 \][/tex]
### Summary
[tex]\[ \begin{aligned} \lim_{x \to 3} [f(x) g(x)] &= -28, \\ \lim_{x \to 3} [3 f(x) g(x)] &= -84, \\ \lim_{x \to 3} [f(x) + 8 g(x)] &= -52, \\ \lim_{x \to 3} \left[\frac{f(x)}{f(x) - g(x)}\right] &\approx 0.36363636363636365. \end{aligned} \][/tex]
Thus, the answers are:
a. [tex]\(-28\)[/tex]
b. [tex]\(-84\)[/tex]
c. [tex]\(-52\)[/tex]
d. [tex]\(\approx 0.3636\)[/tex]
Given:
[tex]\[ \lim_{x \to 3} f(x) = 4 \quad \text{and} \quad \lim_{x \to 3} g(x) = -7 \][/tex]
### Part (a)
[tex]\[ \lim_{x \to 3} [f(x) g(x)] \][/tex]
Using the limit property that states [tex]\(\lim_{x \to a} [f(x) \cdot g(x)] = \left(\lim_{x \to a} f(x)\right) \cdot \left(\lim_{x \to a} g(x)\right)\)[/tex], we have:
[tex]\[ \lim_{x \to 3} [f(x) g(x)] = \left(\lim_{x \to 3} f(x)\right) \cdot \left(\lim_{x \to 3} g(x)\right) = 4 \cdot (-7) = -28 \][/tex]
So,
[tex]\[ \lim_{x \to 3} [f(x) g(x)] = -28 \][/tex]
### Part (b)
[tex]\[ \lim_{x \to 3} [3 f(x) g(x)] \][/tex]
We can factor out the constant 3 from the limit:
[tex]\[ \lim_{x \to 3} [3 f(x) g(x)] = 3 \cdot \lim_{x \to 3} [f(x) g(x)] \][/tex]
From part (a), we know:
[tex]\[ \lim_{x \to 3} [f(x) g(x)] = -28 \][/tex]
So,
[tex]\[ \lim_{x \to 3} [3 f(x) g(x)] = 3 \cdot (-28) = -84 \][/tex]
### Part (c)
[tex]\[ \lim_{x \to 3} [f(x) + 8 g(x)] \][/tex]
Using the limit property that states [tex]\(\lim_{x \to a} [f(x) + g(x)] = \left(\lim_{x \to a} f(x)\right) + \left(\lim_{x \to a} g(x)\right)\)[/tex], we have:
[tex]\[ \lim_{x \to 3} [f(x) + 8 g(x)] = \lim_{x \to 3} f(x) + 8 \cdot \lim_{x \to 3} g(x) \][/tex]
Given,
[tex]\[ \lim_{x \to 3} f(x) = 4 \quad \text{and} \quad \lim_{x \to 3} g(x) = -7 \][/tex]
So,
[tex]\[ \lim_{x \to 3} [f(x) + 8 g(x)] = 4 + 8 \cdot (-7) = 4 + (-56) = 4 - 56 = -52 \][/tex]
### Part (d)
[tex]\[ \lim_{x \to 3} \left[\frac{f(x)}{f(x) - g(x)}\right] \][/tex]
Using the limit properties and given:
[tex]\[ \lim_{x \to 3} f(x) = 4 \quad \text{and} \quad \lim_{x \to 3} g(x) = -7 \][/tex]
First, we find the limit of the denominator:
[tex]\[ \lim_{x \to 3} [f(x) - g(x)] = \lim_{x \to 3} f(x) - \lim_{x \to 3} g(x) = 4 - (-7) = 4 + 7 = 11 \][/tex]
Since the denominator does not approach zero, we can use the limit properties for division:
[tex]\[ \lim_{x \to 3} \left[\frac{f(x)}{f(x) - g(x)}\right] = \frac{\lim_{x \to 3} f(x)}{\lim_{x \to 3} [f(x) - g(x)]} = \frac{4}{11} \][/tex]
So, we have:
[tex]\[ \lim_{x \to 3} \left[\frac{f(x)}{f(x) - g(x)}\right] = \frac{4}{11} \approx 0.36363636363636365 \][/tex]
### Summary
[tex]\[ \begin{aligned} \lim_{x \to 3} [f(x) g(x)] &= -28, \\ \lim_{x \to 3} [3 f(x) g(x)] &= -84, \\ \lim_{x \to 3} [f(x) + 8 g(x)] &= -52, \\ \lim_{x \to 3} \left[\frac{f(x)}{f(x) - g(x)}\right] &\approx 0.36363636363636365. \end{aligned} \][/tex]
Thus, the answers are:
a. [tex]\(-28\)[/tex]
b. [tex]\(-84\)[/tex]
c. [tex]\(-52\)[/tex]
d. [tex]\(\approx 0.3636\)[/tex]
