lim(x趋于0)(tanx-sinx)/3x^2
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lim(x趋于0)(tanx-sinx)/3x^2
lim(x趋于0)(tanx-sinx)/3x^2
lim(x趋于0)(tanx-sinx)/3x^2
0/0型,用洛比达法则,
lim(x趋于0)(tanx-sinx)/(3x^2)=lim(x趋于0)(tanx-sinx)'/(3x^2)'=lim(x趋于0)[(tanx)'-(sinx)']/(6x)
=lim(x趋于0)[1/(cosx)^2-cosx]/(6x)==lim(x趋于0)[1/(cosx)^2-cosx]'/(6x)'
=lim(x趋于0){[1/(cosx)^2]'-(cosx)'}/6=lim(x趋于0)[(2sinx)/(cosx)^3+sinx]/6=[(2sin0)/(cos0)^3+sin0]/6=0
(tanx-sinx)/3x^2~x(1/cosx-1)/3x^2
~(1-cosx)/3xcosx~2[sin(x/2)]^2/3xcosx=0
0
那我就不用洛必达法则了,用定理lim[x→0] sinx/x=1
lim[x→0] (tanx-sinx)/3x²
= (1/3)lim[x→0] (sinx/cosx-sinx)/x²
= (1/3)lim[x→0] (sinx-sinxcosx)/(x²cosx)
= (1/3)lim[x→0] sinx(1-cosx)/(x...
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那我就不用洛必达法则了,用定理lim[x→0] sinx/x=1
lim[x→0] (tanx-sinx)/3x²
= (1/3)lim[x→0] (sinx/cosx-sinx)/x²
= (1/3)lim[x→0] (sinx-sinxcosx)/(x²cosx)
= (1/3)lim[x→0] sinx(1-cosx)/(x²cosx)
= (1/3)lim[x→0] sin²x(1-cosx)/(x²sinxcosx)
= (1/3)lim[x→0] (sinx/x)²·(1-cosx)/(sinxcosx)
= (1/3)lim[x→0] (sinx/x)²·[2sin²(x/2)]/[2sin(x/2)cos(x/2)cosx]
= (1/3)lim[x→0] (sinx/x)²·sin(x/2)/[cos(x/2)cosx]
= (1/3)·1·0/(1·1)
= 0
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