Limits
Evaluate the value of the limit \lim_{x \to 0} \frac{\tan x - \sin x}{x^3}, which arises frequently in expansion-based JEE problems.
Select the correct option:
Solution
21
The fastest reliable route is Taylor series, since both \tan x and \sin x agree to first order, forcing the leading behaviour into the cubic term. Using \sin x = x - x^3/6 + \cdots and \tan x = x + x^3/3 + \cdots, the numerator becomes (x + x^3/3) - (x - x^3/6) = x^3(1/3 + 1/6) = x^3(1/2). Dividing by x^3 gives the limit 1/2. A complementary factoring view writes \tan x - \sin x = \sin x(1 - \cos x)/\cos x \approx x \cdot (x^2/2)/1, again giving 1/2. Option 0 results from stopping at first-order expansion where the linear terms cancel, hiding the true cubic contribution. Option 1/6 isolates only the sine cubic term and ignores the tangent term. Option 1/3 isolates only the tangent term and drops the sine correction. The governing pattern is the standard JEE Advanced higher-order trigonometric limit resolved by matching cubic coefficients. Plausibility check: both \tan x - \sin x and x^3 are positive for small positive x, so the limit must be positive, and the magnitude 1/2 sits sensibly between the two partial contributions 1/3 and 1/6, confirming the answer.
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About This Question
- Subject
- mathematics
- Chapter
- limit, continuity and differentiability
- Topic
- limits
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
21
The fastest reliable route is Taylor series, since both \tan x and \sin x agree to first order, forcing the leading behaviour into the cubic term. Using \sin x = x - x^3/6 + \cdots and \tan x = x + x^3/3 + \cdots, the numerator becomes (x + x^3/3) - (x - x^3/6) = x^3(1/3 + 1/6) = x^3(1/2). Dividing by x^3 gives the limit 1/2. A complementary factoring view writes \tan x - \sin x = \sin x(1 - \cos x)/\cos x \approx x \cdot (x^2/2)/1, again giving 1/2. Option 0 results from stopping at first-order expansion where the linear terms cancel, hiding the true cubic contribution. Option 1/6 isolates only the sine cubic term and ignores the tangent term. Option 1/3 isolates only the tangent term and drops the sine correction. The governing pattern is the standard JEE Advanced higher-order trigonometric limit resolved by matching cubic coefficients. Plausibility check: both \tan x - \sin x and x^3 are positive for small positive x, so the limit must be positive, and the magnitude 1/2 sits sensibly between the two partial contributions 1/3 and 1/6, confirming the answer.
This medium difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of limits. It appeared in the 2025 exam.
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