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To solve the given limit using continuity, we have the following problem:

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

### Step 1: Check continuity
The square root function \( \sqrt{f(x)} \) is continuous wherever its argument \( f(x) \) is non-negative and continuous. So, we need to evaluate the expression inside the square root:

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

### Step 2: Direct substitution
Since \( f(x) \) is a polynomial, it is continuous for all real values of \( x \). Thus, we can directly substitute \( x = 4 \) into the function.

Substituting \( x = 4 \) into \( f(x) \):

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

### Step 3: Evaluate the square root
Now that we have \( f(4) = 4 \), we can evaluate the square root:

\[
\sqrt{f(4)} = \sqrt{4} = 2
\]

Thus, the value of the limit is:

\[
\boxed{2}
\]

Let me know if you need any further details or have any questions!

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Here are 5 related questions:
1. What is the general rule for finding limits using continuity?
2. How do we handle limits involving square root functions where the expression inside might be negative?
3. Can you give an example where substitution into a limit results in an indeterminate form?
4. How does the continuity of polynomials help in limit evaluation?
5. What happens if the expression inside the square root had been a complex number?

**Tip:** When using direct substitution for limits, always ensure that the function is continuous at the point of interest.

Use continuity to evaluate the limit: lim (x -> 4) sqrt(x^2 - 2x - 4)

Learn how to evaluate the limit of the square root function sqrt(x^2 - 2x - 4) as x approaches 4 using continuity principles. This detailed solution explains how to apply the continuity of polynomials and square root functions.

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