limx^2 y^2(x,y)趋于0的极限

lim(n->∝)√n*√f(2/n)=lim(n->∝)√2*√[f(2/n)/(2/n)]=√2lim(n->∝)√f(2/n)/(2/n)n->∝,2/n->0,u=2/n=√2lim(u->0

1,函数在x0处有定义2,在x0处既有左极限又有右极限,且左极限等于右极限3,极限值等于函数值

根号（x平方+2x）+x=[(x平方+2x)-x平方]/[(根号x平方+2x)-x]=2x/[(根号x平方+2x)-x]=2/[(根号1+2/x)-1]当x趋于-∞时,根号1+2/x趋于1,∴(根号1

(x-y)(z+y-x+y+2y)÷4y=(x-y)(z-x+4y)÷4y{(x+y)(x-y)-(x-y)的2次方+2y(x-y)}除以4y=(x-y)(x+y-x+y+2y)÷4y=(x-y)(4

1.∵x²+y²≥2|xy|∴0≤|(x+y)/(x²+y²-xy)|=|x+y|/|x²+y²-xy|≤|x+y|/(x²+y&

由limx→0，y→0f(x，y)-xy(x2+y2)2=1知，因此分母的极限趋于0，故分子的极限必为零，从而有f（0，0）=0；因为极限等于1；故f（x，y）-xy～（x2+y2）2（|x|，|y|

(x^2-y^2)/(x+y)-(4x(x-y)+y^2)/(2x-y)=(x-y)(x+y)/(x+y)-(4x^2-4xy+y^2)/(2x-y)=(x-y)-(2x-y)^2/(2x-y)=(x

1)x(x-y)(x+y)-x(x+y)^2=x((x-y)(x+y)-(x+y)^2)=x(x^2-y^2-x^2-2xy-y^2)=x(-2xy-2y^2)=-2xy(x+y)2)（2a+b)(2

【(x-y)^2+(x+y)(x-y)】除以2x=(x-y)*(x-y+x+y)/2x=(x-y)*2x/2x=x-y

[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y=[x²-y²-x²-2xy-y²-xy+xy²]/(y/2)=[(x-2)y

x→0,y→0时,√x²+y²→0不妨设√x²+y²=t原式=lim(t→0)(t-sint)/t³用洛必达法则得极限值为1/6不明白追问吧~

直接将x=2y=0代入其中,得ln(2+e*0)/根号2*2+0*2)=ln3/2

(x-3y)(x+y)-(x-2y)(x+2y)-(x-y)平方=x²-2xy-3y²-x²+4y²-x²+2xy-y²=(x²-

=(x-y)[(3x+y)-(2x+3y)]*(x-y)=(x-y)*(x-y)(3x+y-2x-3y)=(x-y)(x-y)(x-2y)到底是化简,还是?=(x*x-2xy+y*y)(x-2y)=x

（1)x^2/x)-y-x-y=x-y-x-y=-2y(2)(a/a-b)-(a/a+b)-(2b^2/a^2-b^2)=a(a+b-a+b)/(a^2-b^2)-(2b^2/a^2-b^2)=2b/

y的值为无穷大limx的极限

解题思路：根据题意，由整式的运算的知识整理，分析可以求得解题过程：

答：x趋于π,分母sin(xπ)趋于0因为：极限存在所以：分子趋于0所以：1+b+c=0分子分母同时求导：(2x+b)/[πcos(πx)]=52+b=-5π解得：b=-5π-2所以：c=5π+1再问

令t=e^x-1,x=ln(t+1)原式=t/ln(t+1)=1/[(1/t)ln(t+1)]=1/ln(1+t)^(1/t)(t->0)=1/lne=1解法2原式=(e^x-1)/x(x->0)=（

原式=(x+y)(x-y)[(x+y)(x-y)-(x²+y²)]=(x+y)(x-y)(x²-y²-x²-y²)=-2y²(x+