Question

(a) Prove that the function \( f(x) = |x - 1| \) is continuous at \( x = 1 \):

Hint:

For continuity, ensure \( f(a) = \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) \).

Answer 1

The function \( f(x) = |x - 1| \) can be written as: \[ f(x) = \begin{cases} x - 1 & \text{if } x \geq 1,
1 - x & \text{if } x<1. \end{cases} \] To check continuity at \( x = 1 \), we evaluate:
1. \( f(1) = |1 - 1| = 0. \)
2. Left-hand limit (\( \lim_{x \to 1^-} f(x) \)): \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1 - x) = 0. \] 3. Right-hand limit (\( \lim_{x \to 1^+} f(x) \)): \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 1) = 0. \] Since \( f(1) = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \), the function is continuous at \( x = 1 \).