求下面两个极限lim x[(x^2+1)^(1/2)-x] x趋近于+∞;lim (tanx-sinx)/x^3 x趋近于0;

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求下面两个极限lim x[(x^2+1)^(1/2)-x] x趋近于+∞;lim (tanx-sinx)/x^3 x趋近于0;
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求下面两个极限lim x[(x^2+1)^(1/2)-x] x趋近于+∞;lim (tanx-sinx)/x^3 x趋近于0;
求下面两个极限
lim x[(x^2+1)^(1/2)-x] x趋近于+∞;
lim (tanx-sinx)/x^3 x趋近于0;

求下面两个极限lim x[(x^2+1)^(1/2)-x] x趋近于+∞;lim (tanx-sinx)/x^3 x趋近于0;
lim x[(x^2+1)^(1/2)-x]=lim x/ [(x^2+1)^(1/2)+x]
用洛必达法则=lim 1/[x/(x^2+1)^(1/2)+1]=
lim (x^2+1)^(1/2)/[(x^2+1)^(1/2)+x]=1-limx/ [(x^2+1)^(1/2)+x]
得到2lim x/ [(x^2+1)^(1/2)+x]=1
即 lim x/ [(x^2+1)^(1/2)+x]=1/2
lim x[(x^2+1)^(1/2)-x]=1/2
lim (tanx-sinx)/x^3=lim sinx(1-cosx)/(x^3*cosx)
=lim sinx*2[sin(x/2)]^2/(x^3*cosx)=
lim sinx/x*lim (sin(x/2))^2/(x/2)^2*lim1/2cosx=1/2

第一个1/2

第一题不太会,sorry
第二题是1/2

limx[(x^2+1)^(1/2)-x]
=lim[x/(√(x²+1)+x)]
=lim[1/(√(1+1/x²)+1]
=1/2.
lim (tanx-sinx)/x³=lim[sinx(1-cosx)]/x³cosx
=lim(1-cosx)/x²=1/2
(sinx∽x,(1-cosx)∽x²/2)

第一题
原式=lim x[(x^2+1)^(1/2)-x)*[(x^2+1)^(1/2)+x]/[(x^2+1)^(1/2)+x] x趋近于+∞
=lim x/[(x^2+1)^(1/2)+x] x趋近于+∞
上下再同时出x得:
=lim 1/{[(x^2+1)^(1/2)+x]/x+1} x趋近于+∞
= 1/2
第二题

全部展开

第一题
原式=lim x[(x^2+1)^(1/2)-x)*[(x^2+1)^(1/2)+x]/[(x^2+1)^(1/2)+x] x趋近于+∞
=lim x/[(x^2+1)^(1/2)+x] x趋近于+∞
上下再同时出x得:
=lim 1/{[(x^2+1)^(1/2)+x]/x+1} x趋近于+∞
= 1/2
第二题
=lim sinx(1-cosx)/(x^3*cosx) x趋近于0
=lim sinx*2[sin(x/2)]^2/(x^3*cosx) x趋近于0
=lim sinx/x*lim (sin(x/2))^2/(x/2)^2*lim1/2cosx x趋近于0
=1/2

收起

lim x[(x^2+1)^(1/2)-x] x
=lim x[x/(x^2+1)^(1/2)+x]
=1/2
lim(tanx-sinx)/x^3
=lim sinx(1-cosx)/cosx x^3
=4lim(sinx2sin^2 x/2)/cosx x (x/2)^2
=4

不用这么麻烦
第一题:原式乘上[(x^2+1)^(1/2)+x]再除去它,然后你应该会了。