Show that `f(x)=|x-3|` is continuous but not differentiable at `x=3` .

Explanation

To show that the function \( f(x) = |x - 3| \) is continuous but not differentiable at \( x = 3 \), we will follow these steps:

### Step 1: Check Continuity at \( x = 3 \)

A function is continuous at a point if the following three conditions are satisfied:
1. \( f(c) \) is defined.
2. \( \lim_{x \to c} f(x) \) exists.
3. \( \lim_{x \to c} f(x) = f(c) \).

Let \( c = 3 \).

1. **Evaluate \( f(3) \)**:
\[ f(3) = |3 - 3| = |0| = 0. \]

2. **Find the left-hand limit as \( x \) approaches 3**:
\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} |x - 3| = \lim_{x \to 3^-} -(x - 3) = \lim_{x \to 3^-} (3 - x) = 0. \]

3. **Find the right-hand limit as \( x \) approaches 3**:
\[ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} |x - 3| = \lim_{x \to 3^+} (x - 3) = 0. \]

4. **Check if the limits are equal and equal to \( f(3) \)**:
\[ \lim_{x \to 3^-} f(x) = 0, \quad \lim_{x \to 3^+} f(x) = 0, \quad f(3) = 0. \]
Since both limits equal \( f(3) \), we conclude that \( f(x) \) is continuous at \( x = 3 \).

### Step 2: Check Differentiability at \( x = 3 \)

A function is differentiable at a point if the left-hand derivative and the right-hand derivative exist and are equal.

1. **Find the left-hand derivative at \( x = 3 \)**:
\[ f'(3^-) = \lim_{h \to 0^-} \frac{f(3 + h) - f(3)}{h} = \lim_{h \to 0^-} \frac{|3 + h - 3| - 0}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} -1 = -1. \]

2. **Find the right-hand derivative at \( x = 3 \)**:
\[ f'(3^+) = \lim_{h \to 0^+} \frac{f(3 + h) - f(3)}{h} = \lim_{h \to 0^+} \frac{|3 + h - 3| - 0}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} 1 = 1. \]

3. **Compare the left-hand and right-hand derivatives**:
\[ f'(3^-) = -1, \quad f'(3^+) = 1. \]
Since \( f'(3^-) \neq f'(3^+) \), the function is not differentiable at \( x = 3 \).

### Conclusion

Thus, we have shown that the function \( f(x) = |x - 3| \) is continuous at \( x = 3 \) but not differentiable at that point.

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