求极限~lim(x→0)[(1+xsinx)^1/2-cosx]/sin^2(x/2)
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求极限~lim(x→0)[(1+xsinx)^1/2-cosx]/sin^2(x/2)
lim(x→0)[(1+xsinx)^1/2-cosx]/sin^2(x/2)
=lim(x→0)[(1+xsinx)^1/2-1+1-cosx]/sin^2(x/2)
=lim(x→0)[(1+xsinx)^1/2-1+2sin^2(x/2)]/sin^2(x/2)
=lim(x→0)[(1+xsinx)^1/2-1]/sin^2(x/2)+2 ([(1+x)^1/n]-1~(1/n)*x )
=lim(x→0)1/2xsinx/sin^2(x/2)+2
=lim1/2x^2/ (x/2)^2+2
=2+2
=4
=lim(x→0)[(1+xsinx)^1/2-1+1-cosx]/sin^2(x/2)
=lim(x→0)[(1+xsinx)^1/2-1+2sin^2(x/2)]/sin^2(x/2)
=lim(x→0)[(1+xsinx)^1/2-1]/sin^2(x/2)+2 ([(1+x)^1/n]-1~(1/n)*x )
=lim(x→0)1/2xsinx/sin^2(x/2)+2
=lim1/2x^2/ (x/2)^2+2
=2+2
=4
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