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Question

lim 2h0 3 sin π/6 + h-cos π/6+h3 h 3 cos h-sin h is equal to

(a) 2/3
(b) 4/3
(c) -23
(d) −4/3

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Solution

(d) 4/3

limh0 23 sin π/6+h-cos π/6+h3 h3cos h-sin h=limh0 232cos h+32 sin h-32cos h+sin h2h3 cos h-3 sin h=limh0 22 sin hh×13 cos h-3sin h=limh0 43 cos h-3 sin h=43

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