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MathsMediumMCQ2014 · 06 Apr

Q72. sin(πcos2x) lim is equal to x→0 x2 (1) −π (2) π (3) π (4) 1 2

What This Question Tests

This question tests the evaluation of limits involving trigonometric functions, requiring algebraic manipulation to apply standard limit forms or the use of L'Hopital's rule.

Concepts Tested

Standard trigonometric limitsL'Hopital's ruleTrigonometric identities

Formulas Used

lim(u→0) sin(u)/u = 1

cos 2x = 1 - 2sin²x

lim(u→0) (1-cos u)/u² = 1/2

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