3 √ 9 x − 3 √ 3 + x − √ 2 x

To solve the limit

[ \lim_{x \to 3} \left( \sqrt[3]{9x} - \sqrt[3]{3 + x} - \sqrt{2x} \right) ]

we'll evaluate each term separately to determine if we can apply direct substitution or if further techniques are required.

Step 1: Substitute $x = 3$ Directly

Let's attempt direct substitution to see if it results in a defined value.

1. First Term: $\sqrt[3]{9x} = \sqrt[3]{9 \times 3} = \sqrt[3]{27} = 3$
2. Second Term: $\sqrt[3]{3 + x} = \sqrt[3]{3 + 3} = \sqrt[3]{6}$
3. Third Term: $\sqrt{2x} = \sqrt{2 \times 3} = \sqrt{6}$

Putting these values into the limit expression, we get:

$3 - \sqrt[3]{6} - \sqrt{6}$

Conclusion

Since each term evaluated individually gives a finite value, the limit is:

[ \lim_{x \to 3} \left( \sqrt[3]{9x} - \sqrt[3]{3 + x} - \sqrt{2x} \right) = 3 - \sqrt[3]{6} - \sqrt{6} ]

Would you like more details, or do you have any questions?

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Related Questions to Explore

1. How would this limit change if $x \to 0$ instead of $x \to 3$?
2. What techniques could help if we encountered an indeterminate form like $\frac{0}{0}$?
3. How do we evaluate limits involving higher roots, like the fourth root or fifth root?
4. What are some applications of limits in real-world scenarios?
5. How can we use L'Hôpital's Rule when there are radical expressions?

Tip: When substituting directly into a limit, always check if it leads to a defined number. If not, try algebraic manipulation or L'Hôpital's Rule.