Question

assume \\( \lim_{x \to 4} f(x) = 6 \\) and \\( \lim_{x \to 4} g(x) = 2 \\). compute the following limit and state the limit laws used to justify the computation. \\( \lim_{x \to 4} \sqrt3{f(x)g(x) + 15} \\) \\( \lim_{x \to 4} \sqrt3{f(x)g(x) + 15} = 3 \\) (simplify your answer.) select each limit law used to justify the computation. \\( \square \\) a. difference \\( \square \\) b. product \\( \square \\) c. constant multiple \\( \square \\) d. quotient \\( \square \\) e. sum \\( \square \\) f. power \\( \square \\) g. root

Explanation:

Step1: Apply Product Law

First, use the Product Law for limits: $\lim_{x\to a}[f(x)g(x)] = \lim_{x\to a}f(x) \cdot \lim_{x\to a}g(x)$. So, $\lim_{x\to 4}[f(x)g(x)] = \lim_{x\to 4}f(x) \cdot \lim_{x\to 4}g(x) = 6 \cdot 2 = 12$.

Step2: Apply Sum Law

Next, use the Sum Law: $\lim_{x\to a}[f(x) + c] = \lim_{x\to a}f(x) + \lim_{x\to a}c$ (where $c$ is a constant). Here, $\lim_{x\to 4}[f(x)g(x) + 15] = \lim_{x\to 4}[f(x)g(x)] + \lim_{x\to 4}15 = 12 + 15 = 27$.

Step3: Apply Root Law

Finally, use the Root Law: $\lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\to a}f(x)}$ (for $n$ odd, or $f(x) \geq 0$ for $n$ even). So, $\lim_{x\to 4}\sqrt[3]{f(x)g(x) + 15} = \sqrt[3]{\lim_{x\to 4}[f(x)g(x) + 15]} = \sqrt[3]{27} = 3$.

The limit laws used are Product (B), Sum (E), and Root (G).

Answer:

The value of the limit is $3$. The limit laws used are B. Product, E. Sum, G. Root.