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Question

limxπ41tanx12sinx

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Solution

As sinπ/4=12,tanπ/4=1 limit is of 00 form,
applying l'opital, limxπ/4sec2xxcosx=limxπ4sec3x
=(2)3=23/2

1179139_1272213_ans_55c5336f4d7c485fa54f6b947bcc6e2f.jpg

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