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CA Foundation · Quantitative Aptitude · Sets, Relations and Functions, Limits and Continuity

Let f(x) = |x − 3|/(x − 3) for x ≠ 3 and f(3) = 1. Which statement about f at x = 3 is correct?

f is discontinuous at x = 3 because its left-hand limit is −1 while its right-hand limit is 1. Since the limit does not exist, defining f(3) = 1 cannot make the function continuous there.

  1. Af is continuous because f(3) is defined
  2. Bf is discontinuous because the left-hand limit is −1 and the right-hand limit is 1Correct
  3. Cf is discontinuous because the limit is 0
  4. Df is continuous because both one-sided limits equal 1

Explanation

For x > 3, |x − 3| = x − 3 so f(x) = 1, giving a right-hand limit of 1. For x < 3, |x − 3| = −(x − 3) so f(x) = −1, giving a left-hand limit of −1. The one-sided limits differ, so the limit does not exist and f cannot be continuous, even though f(3) is defined.

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