Show that `f(x) = |x-5|` is continuous but not differentiable at `x = 5`.

Explanation

To show that the function \( f(x) = |x - 5| \) is continuous but not differentiable at \( x = 5 \), we will follow these steps:

### Step 1: Define the function in piecewise form
The absolute value function can be expressed in a piecewise manner:
\[ f(x) = \begin{cases} x - 5 & \text{if } x \geq 5 \\ -(x - 5) = -x + 5 & \text{if } x < 5 \end{cases} \]

### Step 2: Check for continuity at \( x = 5 \)
To show that \( f(x) \) is continuous at \( x = 5 \), we need to verify that:
\[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5) \]

- **Calculate \( f(5) \)**:
\[ f(5) = |5 - 5| = |0| = 0 \]

- **Calculate the left-hand limit \( \lim_{x \to 5^-} f(x) \)**:
For \( x < 5 \):
\[ f(x) = -x + 5 \]
Thus,
\[ \lim_{x \to 5^-} f(x) = -5 + 5 = 0 \]

- **Calculate the right-hand limit \( \lim_{x \to 5^+} f(x) \)**:
For \( x \geq 5 \):
\[ f(x) = x - 5 \]
Thus,
\[ \lim_{x \to 5^+} f(x) = 5 - 5 = 0 \]

Since both limits equal \( f(5) \):
\[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5) = 0 \]
This shows that \( f(x) \) is continuous at \( x = 5 \).

### Step 3: Check for differentiability at \( x = 5 \)
To check for differentiability, we need to find the left-hand derivative and the right-hand derivative at \( x = 5 \).

- **Calculate the left-hand derivative**:
\[ f'(x) = \frac{d}{dx}(-x + 5) = -1 \quad \text{for } x < 5 \]
Thus,
\[ \lim_{x \to 5^-} f'(x) = -1 \]

- **Calculate the right-hand derivative**:
\[ f'(x) = \frac{d}{dx}(x - 5) = 1 \quad \text{for } x \geq 5 \]
Thus,
\[ \lim_{x \to 5^+} f'(x) = 1 \]

Since the left-hand derivative and the right-hand derivative at \( x = 5 \) are not equal:
\[ \lim_{x \to 5^-} f'(x) = -1 \quad \text{and} \quad \lim_{x \to 5^+} f'(x) = 1 \]
This shows that \( f(x) \) is not differentiable at \( x = 5 \).

### Conclusion
We have shown that \( f(x) = |x - 5| \) is continuous at \( x = 5 \) but not differentiable at that point.

---