Discuss the continuity of the function of given by `f(x)=|x-1|+|x-2| at x=1 and x=2`

Explanation

To discuss the continuity of the function \( f(x) = |x - 1| + |x - 2| \) at \( x = 1 \) and \( x = 2 \), we need to check the left-hand limit (LHL), right-hand limit (RHL), and the function value at these points.

### Step 1: Check continuity at \( x = 1 \)

1. **Calculate \( f(1) \)**:
\[ f(1) = |1 - 1| + |1 - 2| = 0 + 1 = 1 \]

2. **Calculate the right-hand limit as \( x \to 1^+ \)**:
\[ f(1 + h) = |(1 + h) - 1| + |(1 + h) - 2| = |h| + |h - 1| \]
For small \( h > 0 \):
\[ f(1 + h) = h + (h - 1) = 2h - 1 \]
Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} f(1 + h) = \lim_{h \to 0} (2h - 1) = -1 \]

3. **Calculate the left-hand limit as \( x \to 1^- \)**:
\[ f(1 - h) = |(1 - h) - 1| + |(1 - h) - 2| = |-h| + |-(1 + h)| = h + (1 + h) = 1 + 2h \]
Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} (1 + 2h) = 1 \]

4. **Conclusion for \( x = 1 \)**:
Since \( \lim_{x \to 1^+} f(x) \neq \lim_{x \to 1^-} f(x) \), the function is not continuous at \( x = 1 \).

### Step 2: Check continuity at \( x = 2 \)

1. **Calculate \( f(2) \)**:
\[ f(2) = |2 - 1| + |2 - 2| = 1 + 0 = 1 \]

2. **Calculate the right-hand limit as \( x \to 2^+ \)**:
\[ f(2 + h) = |(2 + h) - 1| + |(2 + h) - 2| = |1 + h| + |h| = 1 + h \]
Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} f(2 + h) = 1 \]

3. **Calculate the left-hand limit as \( x \to 2^- \)**:
\[ f(2 - h) = |(2 - h) - 1| + |(2 - h) - 2| = |1 - h| + |-(h)| = (1 - h) + 0 = 1 - h \]
Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} f(2 - h) = 1 \]

4. **Conclusion for \( x = 2 \)**:
Since \( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^-} f(x) = f(2) = 1 \), the function is continuous at \( x = 2 \).

### Final Conclusion:
The function \( f(x) = |x - 1| + |x - 2| \) is continuous at \( x = 2 \) but not continuous at \( x = 1 \).

---