Question

Prove that the function \( f(x) = |x| \), is continuous at \( x = 0 \).

Hint:

{Key Points:}

• For continuity at a point, LHL = RHL = f(a)
• For \( |x| \), LHL = RHL = 0 at x = 0
• The function is continuous but not differentiable at x = 0 (sharp corner)

Answer 1

We need to prove that the function \( f(x) = |x| \) is continuous at \( x = 0 \). Definition of Continuity: A function \( f(x) \) is said to be continuous at \( x = a \) if: \[ \lim_{x \to a} f(x) = f(a) \] This requires three conditions:

• \( f(a) \) is defined
• \( \lim_{x \to a} f(x) \) exists
• \( \lim_{x \to a} f(x) = f(a) \)

Step 1: Check if \( f(0) \) is defined \[ f(0) = |0| = 0 \] Thus, \( f(0) \) is defined and equals 0.
Step 2: Find the left-hand limit (LHL) as \( x \to 0^- \) When \( x<0 \), \( |x| = -x \) \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} |x| = \lim_{x \to 0^-} (-x) = 0 \]
Step 3: Find the right-hand limit (RHL) as \( x \to 0^+ \) When \( x>0 \), \( |x| = x \) \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} |x| = \lim_{x \to 0^+} x = 0 \]
Step 4: Compare the limits \[ \text{LHL} = \lim_{x \to 0^-} f(x) = 0 \] \[ \text{RHL} = \lim_{x \to 0^+} f(x) = 0 \] Since LHL = RHL = 0, the limit exists and: \[ \lim_{x \to 0} f(x) = 0 \]
Step 5: Verify the continuity condition \[ \lim_{x \to 0} f(x) = 0 \quad \text{and} \quad f(0) = 0 \] Therefore: \[ \lim_{x \to 0} f(x) = f(0) \] Conclusion: Since all three conditions of continuity are satisfied, the function \( f(x) = |x| \) is **continuous at \( x = 0 \)**. Graphical Interpretation: The graph of \( f(x) = |x| \) is a V-shaped curve. At \( x = 0 \), there is no break or jump in the graph; it is a smooth meeting point of the two lines \( y = -x \) (for \( x<0 \)) and \( y = x \) (for \( x>0 \)).