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Question Number 39635 by math khazana by abdo last updated on 09/Jul/18
calculatelimx→11+cos(πx)x2−sin(πx2)
Commented by abdo mathsup 649 cc last updated on 09/Jul/18
changementx−1=tgivet→0whenx→1and1+cos(πx)x2−sin(πx2)=1+cos(π(1+t))(1+t)2−sin(π(1+t)2)=1−cost1+t2+2t−cos(πt2)=A(t)but1−cost∼t22and1−cos(πt2)∼π2t28(t→0)⇒A(t)∼t22t2+2t+π2t28=t2(t+2+π2t8)t→0→0⇒limx→11+cos(πx)x2−sin(πx2)=0
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jul/18
t=1−xlimt→01+cos{Π(1−t)}(1−t)2−sin{Π(1−t)2}limt→01−cos(Πt)(1−t)2−cosΠt2limt→02sin2Πt21−2t+t2−cosΠt2limt→02sin2Πt22sin2Πt2+t2−2tlimt→02×sin(Πt2)×sin(Πt2)Πt2×Πt2×∏242×sin(Πt2)×sin(Πt2)Πt2×Πt2×∏24+1−2tlimt→0∏22∏22+1−∞=0plscheck...
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