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NDA 2024 Maths · Mathematics · Limits · hard

Since sin (1/x) is bounded but oscillates wildly near the origin, what happens to the quotient (x^2sin (1/x))/(sin x) as x tends to 0? Its limit is

Single correct option

Since sin⁡(1/x)\sin(1/x) is bounded but oscillates wildly near the origin, what happens to the quotient x2sin⁡(1/x)sin⁡x\dfrac{x^2\sin(1/x)}{\sin x} as xx tends to 00? Its limit is
  1. A. 00
  2. B. 11
  3. C. does not exist
  4. D. ∞\infty
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Answer: (A) 00

Write the expression as xsin⁡x⋅xsin⁡1x\frac{x}{\sin x}\cdot x\sin\frac1x. As x→0x\to0, xsin⁡x→1\frac{x}{\sin x}\to1, while ∣xsin⁡1x∣≤∣x∣→0\left|x\sin\frac1x\right|\le|x|\to0 by the sandwich theorem. So the limit is 0. Trap: the factor sin⁡1x\sin\frac1x has no limit on its own, but it is bounded and multiplied by a term tending to zero, so the product has a limit.
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