lim((sqr(2)-sqr(1+cosx))/((sinx)^2)) x->0

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lim((sqr(2)-sqr(1+cosx))/((sinx)^2)) x->0
lim((sqr(2)-sqr(1+cosx))/((sinx)^2)) x->0

lim((sqr(2)-sqr(1+cosx))/((sinx)^2)) x->0
求极限 x→0lim[(√2)-√(1+cosx)]/sin²x
x→0lim[(√2)-√(1+cosx)]/sin²x=x→0lim[2-(1+cosx)]/{[(√2)+√(1+cosx)]sin²x}
=x→0lim(1-cosx)/{[(√2)+√(1+cosx)]sin²x}=x→0lim(1-cosx)/{[(√2)+√(1+cosx)](1-cos²x)}
=x→0lim(1-cosx)/{[(√2)+√(1+cosx)](1-cosx)(1+cosx)}=x→0lim{1/[(√2)+√(1+cosx)](1+cosx)]}
=1/[4(√2)]=(√2)/8.

由cosx=2cos²0.5x-1可得
lim(x→0)(√2-√(1+cosx))/(sin²x)=lim(x→0)√2(1-cos0.5x)/sin²x=lim(x→0)2√2sin²0.25x/sin²x=√2/8

=负4分之根号2

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