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Limits Continuity and Differentiability question

2021 · 17 Mar · Shift 2 · Q29
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  5. /2021 · 17 Mar · Shift 2 · Q29

Limits Continuity and Differentiability question

2021 · 17 Mar · Shift 2 · Q29

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
The value of the limit lim⁡θ→0tan⁡(πcos⁡2θ)sin⁡(2πsin⁡2θ)\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}θ→0lim​sin(2πsin2θ)tan(πcos2θ)​ is equal to :
  1. A
    0
  2. B
    −12-{1 \over 2}−21​
  3. C
    14{1 \over 4}41​
  4. D
    −14-{1 \over 4}−41​
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