Respuesta :
Answer:
[tex]\lim_{x \to 3} \sqrt[3]{f(x)g(x)+10}=4[/tex]
Step-by-step explanation:
We know that:
[tex]\lim_{x \to 3}f(x)=6[/tex]
[tex]\lim_{x \to 3}g(x)=9[/tex]
To find [tex]\lim_{x \to 3} \sqrt[3]{f(x)g(x)+10}[/tex] you must:
Step 1: Use the following definition [tex]\sqrt[n]{x}={{x}^{{\frac{1}{n}}}}[/tex]
[tex]\lim_{x \to 3} \sqrt[3]{f(x)g(x)+10}=\\\\\lim_{x \to 3}\:(f(x)g(x)+10)^{\frac{1}{3} }[/tex]
Step 2: From the Limit Laws use the Power law [tex]\lim_{x \to a} (f(x))^n=( \lim_{x \to a} f(x))^n[/tex]
[tex](\lim_{x \to 3}(f(x)g(x)+10))^{\frac{1}{3}}[/tex]
Step 3: From the Limit Laws use the Addition Law [tex]\lim_{x \to a} f(x)+g(x)=\lim_{x \to a} f(x)+\lim_{x \to a} g(x)[/tex]
[tex](\lim_{x \to 3}f(x)g(x)+\lim_{x \to 3}10)^{\frac{1}{3}}[/tex]
Step 4: From the Limit Laws use the Multiplication Law [tex]\lim_{x \to a} f(x)\cdot g(x)=\lim_{x \to a} f(x)\cdot \lim_{x \to a} g(x)[/tex]
[tex](\lim_{x \to 3}f(x)\cdot \lim_{x \to 3}g(x)+\lim_{x \to 3}10)^{\frac{1}{3}}[/tex]
Step 5: Evaluate the limit
[tex](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[/tex]
