Question:medium

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

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{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)
Updated On: Mar 7, 2026
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Solution and Explanation

To Prove:
We are tasked with proving that the function \( f(x) = |x| \) is continuous at \( x = 0 \). A function is continuous at a point \( x = a \) if the following three conditions are satisfied:
1. \( f(a) \) is defined.
2. \( \lim_{x \to a} f(x) \) exists.
3. \( \lim_{x \to a} f(x) = f(a) \).

Step 1: Check if \( f(0) \) is defined
For \( f(x) = |x| \), the value of the function at \( x = 0 \) is
\[ f(0) = |0| = 0 \] So, \( f(0) \) is defined and equal to 0.

Step 2: Check if the limit \( \lim_{x \to 0} f(x) \) exists
We need to check the left-hand limit and the right-hand limit as \( x \to 0 \).

- For \( x \to 0^- \) (approaching 0 from the left), when \( x \) is negative, \( |x| = -x \). Hence, the limit is:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} |x| = \lim_{x \to 0^-} -x = 0 \] - For \( x \to 0^+ \) (approaching 0 from the right), when \( x \) is positive, \( |x| = x \). Hence, the limit is:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} |x| = \lim_{x \to 0^+} x = 0 \] Since both the left-hand limit and the right-hand limit are equal to 0, we have:
\[ \lim_{x \to 0} f(x) = 0 \] Hence, the limit exists and is equal to 0.

Step 3: Check if \( \lim_{x \to 0} f(x) = f(0) \)
From step 1, we know that \( f(0) = 0 \), and from step 2, we know that \( \lim_{x \to 0} f(x) = 0 \). Since the limit equals the function value at \( x = 0 \), the function satisfies the third condition for continuity.

Conclusion
Since all three conditions for continuity are satisfied, we can conclude that the function \( f(x) = |x| \) is continuous at \( x = 0 \).
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