求极限,高数
3个回答
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let
y=x-1
y->0
cosy ~ 1-(1/2)y^2
ln[cosy] ~ ln[1-(1/2)y^2] ~ -(1/2)y^2
lim(x->1) ln[cos(x-1)]/[ 1- (sin(πx/2))^2 ]
=lim(y->0) ln[cosy]/[ 1- (cos(πy/2))^2 ]
=lim(y->0) ln[cosy]/[ (sin(πy/2))^2 ]
= lim(y->0) -(1/2)y^2/(πy/2)^2
=-2/π^2
y=x-1
y->0
cosy ~ 1-(1/2)y^2
ln[cosy] ~ ln[1-(1/2)y^2] ~ -(1/2)y^2
lim(x->1) ln[cos(x-1)]/[ 1- (sin(πx/2))^2 ]
=lim(y->0) ln[cosy]/[ 1- (cos(πy/2))^2 ]
=lim(y->0) ln[cosy]/[ (sin(πy/2))^2 ]
= lim(y->0) -(1/2)y^2/(πy/2)^2
=-2/π^2
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