求极限:lim(x→0)ln(1+x²)/ (sec x- cos x)
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求极限:lim(x→0)ln(1+x²)/ (sec x- cos x)
求极限:lim(x→0)ln(1+x²)/ (sec x- cos x)
求极限:lim(x→0)ln(1+x²)/ (sec x- cos x)
lim(x→0)ln(1+x²)/ (sec x- cos x)
=lim(x→0)cosx*ln(1+x²)/ sin² x
=lim(x→0)cosx*lim(x→0)ln(1+x²)/ sin² x
=lim(x→0)[x/(1+x²)]/2sinxcosx
=lim(x→0)1/[2(1+x²)cosx]*lim(x→0)x/sinx
=1/2