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Question Number 195325 by mathlove last updated on 30/Jul/23
provethatlimx→π2tan(x2)−1x−π2=1
Answered by BaliramKumar last updated on 30/Jul/23
limx→π2ddx(tan(x2)−1)ddx(x−π2)=sec2(x2)⋅12112sec2(π4)=12(2)2=1
Answered by som(math1967) last updated on 30/Jul/23
limx→π2sinx2−cosx2cosx2(x−π2)limx→π22(12sinx2−12cosx2)2cosx2(x2−π4)limx→π22sin(x2−π4)2×12(x2−π4)limx→π2sin(x2−π4)(x2−π4)let(x2−π4)=tx→π2⇒x2→π4∴(x2−π4)→0∴limt→0sintt=1
Commented by mathlove last updated on 30/Jul/23
tnks
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