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JEE Main 2022 · Mathematics · Limits · hard

Lim x→0((sin x)/(x))^1/x^2 equals

Single correct option

lim⁡x→0(sin⁡xx)1/x2\displaystyle\lim_{x\to0}\left(\dfrac{\sin x}{x}\right)^{1/x^2} equals
  1. A. e1/6e^{1/6}
  2. B. e−1/6e^{-1/6}
  3. C. e−1/3e^{-1/3}
  4. D. 11
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Answer: (B) e−1/6e^{-1/6}

The limit has the form 1∞1^\infty, so take logarithms: L=lim⁡1x2ln⁡sin⁡xxL=\lim \frac{1}{x^2}\ln\frac{\sin x}{x}. Using sin⁡xx=1−x26+x4120−⋯\frac{\sin x}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots and ln⁡(1+u)≈u\ln(1+u)\approx u for small uu, we get ln⁡sin⁡xx=−x26+O(x4)\ln\frac{\sin x}{x}=-\frac{x^2}{6}+O(x^4). Hence L=−16L=-\frac16 and the required limit is e−1/6e^{-1/6}. Because sin⁡x<x\sin x<x, the base is below 11, so a value less than 11 is expected, ruling out e1/6e^{1/6} immediately.
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