limx^2 y^2(x,y)趋于0的极限
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![limx^2 y^2(x,y)趋于0的极限](/uploads/image/f/667125-45-5.jpg?t=limx%5E2+y%5E2%28x%2Cy%29%E8%B6%8B%E4%BA%8E0%E7%9A%84%E6%9E%81%E9%99%90)
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+