Questions: Assume the limit of f(x) as x approaches 6 is 9 and the limit of g(x) as x approaches 6 is 9. Compute the following limit and state the limit laws used to justify the computation. The limit of the cube root of f(x)g(x) - 17 as x approaches 6 equals (Simplify your answer.)

Assume the limit of f(x) as x approaches 6 is 9 and the limit of g(x) as x approaches 6 is 9. Compute the following limit and state the limit laws used to justify the computation.

The limit of the cube root of f(x)g(x) - 17 as x approaches 6 equals (Simplify your answer.)

Transcript text: Assume $\lim _{x \rightarrow 6} f(x)=9$ and $\lim _{x \rightarrow 6} g(x)=9$. Compute the following limit and state the limit laws used to justify the computation. \[ \lim _{x \rightarrow 6} \sqrt[3]{f(x) g(x)-17} \] $\lim _{x \rightarrow 6} \sqrt[3]{f(x) g(x)-17}=$ $\square$ (Simplify your answer.)

Solution

Compute the limit $ \lim_{x \rightarrow 6} \sqrt[3]{f(x) g(x) - 17} $.

Apply the product law of limits.

Using the given limits, we have $ \lim_{x \rightarrow 6} f(x) = 9 $ and $ \lim_{x \rightarrow 6} g(x) = 9 $. Therefore, $ \lim_{x \rightarrow 6} f(x) g(x) = 9 \cdot 9 = 81 $.

Subtract 17 from the limit of the product.

Now, we compute $ 81 - 17 = 64 $.

Apply the cube root to the result.

Finally, we find $ \sqrt[3]{64} = 4 $.

The limit is $ \boxed{4} $.

The limit $ \lim_{x \rightarrow 6} \sqrt[3]{f(x) g(x) - 17} $ is $ \boxed{4} $.

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