几道大学微积分求极限的题
1,lim(x→∞){(x-3)^12(2x+1)^8/(3x-1)^20}2,lim(x→+∞){(3x³+5x²+4)/[(√x^6)+2]}3,...
1,lim(x→∞){(x-3)^12(2x+1)^8/(3x-1)^20}
2,lim(x→+∞){(3x³+5x²+4)/[(√x^6)+2]}
3,lim(x→∞)[x+³√(1-x³)]
4,lim(x→1){[√(x+3)-2]/√x-1}
要过程,不要答案,按情况适当加分 展开
2,lim(x→+∞){(3x³+5x²+4)/[(√x^6)+2]}
3,lim(x→∞)[x+³√(1-x³)]
4,lim(x→1){[√(x+3)-2]/√x-1}
要过程,不要答案,按情况适当加分 展开
1个回答
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lim(x→∞){(x-3)^12(2x+1)^8/(3x-1)^20}
=lim(x→∞){(1-3/x)^12(2+1/x)^8/(3-1/x)^20}
=2^8/3^20
lim(x→+∞){(3x³+5x²+4)/[(√x^6)+2]}
=lim(x→+∞){(3x³+5x²+4)/[(x^3)+2]}
=3
lim(x→∞)[x+³√(1-x³)]
=lim(x→∞)[x+³√(1-x³)][x^2-x³√(1-x³)+³√(1-x³)^2]/[x^2-x³√(1-x³)+³√(1-x³)^2]
=lim(x→∞)1/[x^2-x³√(1-x³)+³√(1-x³)^2]
=0
lim(x→1){[√(x+3)-2]/√x-1}
=lim(x→1){[√(x+3)-2][√(x+3)+2]/[√x-1][√(x+3)+2]}
=lim(x→1)[√x-1]/[√(x+3)+2]
=0
=lim(x→∞){(1-3/x)^12(2+1/x)^8/(3-1/x)^20}
=2^8/3^20
lim(x→+∞){(3x³+5x²+4)/[(√x^6)+2]}
=lim(x→+∞){(3x³+5x²+4)/[(x^3)+2]}
=3
lim(x→∞)[x+³√(1-x³)]
=lim(x→∞)[x+³√(1-x³)][x^2-x³√(1-x³)+³√(1-x³)^2]/[x^2-x³√(1-x³)+³√(1-x³)^2]
=lim(x→∞)1/[x^2-x³√(1-x³)+³√(1-x³)^2]
=0
lim(x→1){[√(x+3)-2]/√x-1}
=lim(x→1){[√(x+3)-2][√(x+3)+2]/[√x-1][√(x+3)+2]}
=lim(x→1)[√x-1]/[√(x+3)+2]
=0
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