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Limits Continuity and Differentiability question

2021 · 31 Aug · Shift 1 · Q35
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  5. /2021 · 31 Aug · Shift 1 · Q35

Limits Continuity and Differentiability question

2021 · 31 Aug · Shift 1 · Q35

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
lim⁡x→0sin⁡2(πcos⁡4x)x4\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}x→0lim​x4sin2(πcos4x)​ is equal to :
  1. A
    π2{\pi ^2}π2
  2. B
    2π22{\pi ^2}2π2
  3. C
    4π24{\pi ^2}4π2
  4. D
    4π4\pi4π
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