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The expression provided is:

\[
\lim_{{x \to 4}} \sqrt[6]{{x^2 - 2x - 4}}
\]

To solve this limit using continuity, we need to evaluate the function directly by substituting \( x = 4 \) into the expression, provided the function is continuous at this point.

1. First, substitute \( x = 4 \) into the polynomial under the root:

\[
f(x) = x^2 - 2x - 4
\]

When \( x = 4 \):

\[
f(4) = 4^2 - 2(4) - 4 = 16 - 8 - 4 = 4
\]

2. Now substitute this value into the root expression:

\[
\sqrt[6]{{f(4)}} = \sqrt[6]{4}
\]

So the limit becomes:

\[
\lim_{{x \to 4}} \sqrt[6]{{4}} = \sqrt[6]{4}
\]

Since there is no indeterminate form, the limit simplifies to \( \sqrt[6]{4} \), which is the sixth root of 4.

Thus, the value of the limit is:

\[
\boxed{\sqrt[6]{4}}
\]

Would you like further explanation or details on how continuity ensures the solution is valid? 

Here are 5 related questions:
1. How can we handle limits that result in indeterminate forms like \( \frac{0}{0} \)?
2. What conditions must be met for a function to be continuous at a point?
3. How do you calculate limits that involve higher-order roots, such as cube or sixth roots?
4. What techniques can be used when substitution alone doesn't work for evaluating limits?
5. How does the concept of continuity help simplify limit problems?

**Tip:** Always check if a function is continuous at a given point by substituting directly. If there’s no indeterminate form, the limit can often be found with basic substitution.

Learn how to evaluate the limit of √[6]{x² - 2x - 4} as x approaches 4 using the continuity theorem. This solution explains key concepts in calculus, including limits and radicals, and is suitable for students in Grades 11-12.

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