求证lim x→∞ √x^2+1-√x^-2x+5 =1
来源：学生作业帮助网 编辑：六六作业网 时间：2024/05/15 18:28:45
求证limx→∞√x^2+1-√x^-2x+5=1求证limx→∞√x^2+1-√x^-2x+5=1求证limx→∞√x^2+1-√x^-2x+5=1分子有理化得原式=lim(x→∞)[(x^2+1)
求证lim x→∞ √x^2+1-√x^-2x+5 =1
求证lim x→∞ √x^2+1-√x^-2x+5 =1
求证lim x→∞ √x^2+1-√x^-2x+5 =1
分子有理化得
原式=lim(x→∞) [(x^2+1)-(x^2-2x+5)]/[√(x^2+1)+√(x^2-2x+5)]
=lim(x→∞) (2x+4)/[√(x^2+1)+√(x^2-2x+5)]
=lim(x→∞) (2x+4)/[√(x^2)+√(x^2)]
=1
求证lim x→∞ √x^2+1-√x^-2x+5 =1
求lim(√x^2+1)-x x→-∞
lim√x(sinx+cosx)/x+1 x→+∞
lim((√1+x^2)+x)/((√x^3+x)-x)x→+无穷大
lim (√(x^2+1)+2x)^2/(3x^2+1) x→＋∞
求极限lim(x→-∞)[√(x^2+x-1)+xe^(1/x)]
计算lim(x →∞) [√(1+x^2)-x]x
lim(√(4x²-8x+5)+2x+1),x→-∞怎么解?
Lim [x^2/(x^2-1)]^x (x→∞)
lim(1+x/x)2x (x→∞)
计算极限① lim(x→+∞) 1 / [ (√x^2+3x) - x ],② lim(x→1) (sinx-sin1)/(x-1)
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,lim(x→∞)[x+³√(1-x³)]4,lim(x→1)｛[√(x+3)-2]/√x-1｝
lim(x →∞) (x√x sin1/x)/(√x -1)
求lim(x->∞):[√(4x^2+x-1)+x+1]/[√(x^2+sinx)]
1.lim x→0 3x/(sinx-x)2.lim x→0 (1-cosmx)/x^23.lim x→0 sin4x/[(√x+2)-√2]4.lim h →0 [sin(x+h)-sin(x-h)]/h
lim √（3-x）-√（1+x）/x平方+x-2x→1
数学极限lim(x→0)((x^3)/(3*x^2-1)) lim(x→+无穷)(√x(x+2)-√(x^2-x+1)) lim(x→1)(tan(x-1))/(x^2-1)
lim√x x→∞和lim√x x→0分别等于多少