To solve for the values of $x$ for which the function is continuous, let's examine the function provided:

$f(x) = \frac{|x - 13|}{x - 13}$

Step 1: Analyze the denominator

The function will be undefined wherever the denominator is zero. So, set the denominator equal to zero:

$x - 13 = 0$ $x = 13$

Thus, the function is undefined at $x = 13$, and we must exclude $x = 13$ from the domain to ensure the function is continuous.

Step 2: Behavior of the function for $x \neq 13$

We need to check what happens to the function for values of $x$ on either side of $x = 13$:

• Case 1: $x > 13$

  If $x > 13$, then $|x - 13| = x - 13$, so the function becomes:

  $f(x) = \frac{x - 13}{x - 13} = 1$

• Case 2: $x < 13$

  If $x < 13$, then $|x - 13| = -(x - 13)$, so the function becomes:

  $f(x) = \frac{-(x - 13)}{x - 13} = -1$

Thus, the function is 1 for all $x > 13$ and -1 for all $x < 13$.

Step 3: Discontinuity at $x = 13$

At $x = 13$, the function has a discontinuity because the function is undefined there and the limits from the left and right do not match:

• As $x \to 13^+$, $f(x) = 1$
• As $x \to 13^-$, $f(x) = -1$

This jump discontinuity at $x = 13$ confirms that the function is not continuous at $x = 13$.

Conclusion

The function is continuous for all $x \neq 13$. The correct answer is:

$\boxed{x \neq 13}$

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

Follow-up Questions:

1. What is the definition of a jump discontinuity in a function?
2. Can the absolute value in the numerator affect the continuity in any other cases?
3. How do you find the limits of a piecewise function like this?
4. Why is the function undefined at $x = 13$?
5. How can we generalize the approach for functions involving absolute values and rational expressions?

Tip:

When analyzing the continuity of a function with absolute values and rational terms, always check the denominator for zeros and investigate behavior around those points carefully.