lim x→5 x-5/x/1-x/5

lim(x→∞)arcsin2x/5xlim(x→∞)arcsin2x/5xlim(x→∞)arcsin2x/5x结果是0,因为式子分子arcsin2x是有界的（-2/π~2/π）,而分母趋近于无穷,

lim(1-(5/x))^x-2lim(1-(5/x))^x-2lim(1-(5/x))^x-2是不是x→∞?令1/a=-5/x则a→∞x=-5ax-2=-5a-2所以原式=(1+1/a)^(-5a-

lim（1-5/x）^xlim（1-5/x）^xlim（1-5/x）^x

lim(x^3+4x+1)/(x^4+5x+4)x→无限lim(x^3+4x+1)/(x^4+5x+4)x→无限lim(x^3+4x+1)/(x^4+5x+4)x→无限lim(x^3+4x+1)/(x

lim(x→∞)[(X^2+2x+5)/(3x^2-x+1)=lim(x→∞)[(X^2+2x+5)/(3x^2-x+1)=lim(x→∞)[(X^2+2x+5)/(3x^2-x+1)=无穷大时只考虑

1.limx->∞（2x/x^2+1)*cosx2..limx->∞{[(x^2-x)arctanx]/x^3-x-5}3..limx->0（e^x-e^-x-2x)/(x-sinx)4..limx-

帮忙`````````一些求极限的数学题一.求极限1.lim3x^+2x+1x→12.limx^+3x+2/x的三次方+4x^+5x→03.limx^-1/x-1x→14.limx^-3x+2/1-x

lim(x→∞)[5X*(tan2/3x)]lim(x→∞)[5X*(tan2/3x)]lim(x→∞)[5X*(tan2/3x)]利用等价无穷小的概念当x取向无穷大：tan2/3x=2/3x所以极限

lim(x→0)(3x^2-5x+2),lim(x→0)(3x^2-5x+2),lim(x→0)(3x^2-5x+2),这不是分式,把0直接代入就行了=2

求极限lim(x→∞)(2x/3-x)-(2/3x)求极限lim(x→5)根号(x-1)-2/x-5求极限lim(x→∞)(2x/3-x)-(2/3x)求极限lim(x→5)根号(x-1)-2/x-5

lim[x→1]f(x)存在,且f(x)=2x+5+3lim[x→1]f(x),求f(x)lim[x→1]f(x)存在,且f(x)=2x+5+3lim[x→1]f(x),求f(x)lim[x→1]f(

烦人的数学微积分1、lim(x→0)Sin3x/Sin2x2、lim(x→0)x/3*Sin3x3、lim(x→0)[Xsin(3/x)+(tanx/2x)]4、lim(x→3)Sin(x-3)/(X

求lim(x→无限)4x^3-1/5x^3-2x+5求lim(x→无限)4x^3-1/5x^3-2x+5求lim(x→无限)4x^3-1/5x^3-2x+5lim(x→无限)4x^3-1/5x^3-2

lim（x^2-1）/(x^2-5x+4)x→1lim（x^2-1）/(x^2-5x+4)x→1lim（x^2-1）/(x^2-5x+4)x→1因式分解，然后把公共的因式（x-1）约分，得到-2/3

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)

lim→1（X^2-3X+2)/(X^2+4X-5)求极限lim→1（X^2-3X+2)/(X^2+4X-5)求极限lim→1（X^2-3X+2)/(X^2+4X-5)求极限lim（x→1)（X^2-

求救~lim(x-1)^10*(2x-3)^10/(3x-5)^20(x→∞)求救~lim(x-1)^10*(2x-3)^10/(3x-5)^20(x→∞)求救~lim(x-1)^10*(2x-3)^

lim3次√(x^3-5x^2+1)-xX→无穷lim3次√(x^3-5x^2+1)-xX→无穷lim3次√(x^3-5x^2+1)-xX→无穷答案不是0,应该是-5/3吧lim(x³-5x

lim(√(4x²-8x+5)+2x+1),x→-∞怎么解?lim(√(4x²-8x+5)+2x+1),x→-∞怎么解?lim(√(4x²-8x+5)+2x+1),x→-

求极限lim(x→∞)[(2x+5)/(2x+3)]^(x+1)求极限lim(x→∞)[(2x+5)/(2x+3)]^(x+1)求极限lim(x→∞)[(2x+5)/(2x+3)]^(x+1)原式=l