Question

2.3.12 Assume lim f(x)-6 and lim g(x)-9.Compute the following limit and state the limit laws used to justify the computation x-+3 lim Vf(x)g(x)+10 x-+3 m Vrx)g)+10(Simplify your answer.) x-+3

Answer 1

Answer:

$\lim_{x \to 3} \sqrt[3]{f(x)g(x)+10}=4$

Step-by-step explanation:

We know that:

$\lim_{x \to 3}f(x)=6$

$\lim_{x \to 3}g(x)=9$

To find  $\lim_{x \to 3} \sqrt[3]{f(x)g(x)+10}$ you must:

Step 1: Use the following definition $\sqrt[n]{x}={{x}^{{\frac{1}{n}}}}$

$\lim_{x \to 3} \sqrt[3]{f(x)g(x)+10}=\\\\\lim_{x \to 3}\:(f(x)g(x)+10)^{\frac{1}{3} }$

Step 2: From the Limit Laws use the Power law $\lim_{x \to a} (f(x))^n=( \lim_{x \to a} f(x))^n$

$(\lim_{x \to 3}(f(x)g(x)+10))^{\frac{1}{3}}$

Step 3: From the Limit Laws use the Addition Law $\lim_{x \to a} f(x)+g(x)=\lim_{x \to a} f(x)+\lim_{x \to a} g(x)$

$(\lim_{x \to 3}f(x)g(x)+\lim_{x \to 3}10)^{\frac{1}{3}}$

Step 4: From the Limit Laws use the Multiplication Law $\lim_{x \to a} f(x)\cdot g(x)=\lim_{x \to a} f(x)\cdot \lim_{x \to a} g(x)$

$(\lim_{x \to 3}f(x)\cdot \lim_{x \to 3}g(x)+\lim_{x \to 3}10)^{\frac{1}{3}}$

Step 5: Evaluate the limit

$(6\cdot 9+\lim_{x \to 3}10)^{\frac{1}{3}}\\\left(6\cdot \:9+10\right)^{\frac{1}{3}}\\64^{\frac{1}{3}}\\\left(4^3\right)^{\frac{1}{3}}\\4$