Continuity
Suppose f(x) = \frac{\sqrt{1+x} - \sqrt{1-x}}{x} for x \neq 0 and f(0) = k; which k makes f continuous at the origin?
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About This Question
- Subject
- mathematics
- Chapter
- limit, continuity and differentiability
- Topic
- continuity
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
1
Continuity at the origin demands that the assigned value k equal the limit of the expression as x \to 0, so the task is to evaluate that limit. Rationalize by multiplying numerator and denominator by \sqrt{1+x} + \sqrt{1-x}: the numerator becomes (1+x) - (1-x) = 2x, giving f(x) = \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2}{\sqrt{1+x} + \sqrt{1-x}}. As x \to 0 the denominator tends to 1 + 1 = 2, so the limit is 2/2 = 1. Hence k = 1. Option 0 wrongly assumes the leading behaviour cancels to zero, but the cancellation is between equal-order roots, not a vanishing one. Option 1/2 stops after halving without completing the rationalization. Option 2 forgets to divide by the denominator sum of the two roots. The governing JEE Advanced pattern is conjugate rationalization to remove a 0/0 surd indeterminacy. Plausibility check: at x = 0.01, f \approx (1.00499 - 0.99499)/0.01 = 1.00003, confirming the limit and the continuity value k = 1.
This easy difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of continuity. It appeared in the 2025 exam.
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