limx→13x+1x-3

limx→-1(x^3/x+1)∵limx→-1(x+1/x^3)=0/(-1)=0而无穷小的倒数=无穷大∴原式=∞.原式=-1/0=-∞

limx→-31/x-3=-1/6

求解limx^(1/1-x)x→1lim((2-x)/(3-x))^(1/x)x→1limx^(1/(1-x))x→1lim((2-x)/(3-x))^(1/x)x→1第二个问题错了lim((2-x)/(3-x))^(1/x)x→0limx

limx→0(tanx-sinx)/sin^3x=limx→0(tanx-sinx)/x³为什么可以直接去掉sinxtanx可以写为sinx/cosx所以可以去掉sinx

limx→3e^1/(x-3)(x-2）这个极限不存在,左极限0,右极限无穷

limx→27((x-27)/(x^(1/3)-3))因为x-27=(x^(1/3)-3))[x^(2/3)+3x^(1/3)+9]所以((x-27)/(x^(1/3)-3))=x^(2/3)+3x^(1/3)+9所以,原极限=limx→2

limx→∞(1+1/2x)^3x+2limx→∞(1+1/2x)^3x+2=limx→∞(1+1/2x)^2x*(3x+2)/(2x)=e^limx→∞(3x+2)/(2x)=e^(3/2)

limx→-1.x^2-1/x^3-1如果分母是x^3-1,直接代入x=-1,极限=0猜测分母是x^3+1因式分解limx→-1.x^2-1/x^3+1=limx→-1(x-1)(x+1)/(x+1)(x^2-x+1)=limx→-1(x-

limx→0(sinx/x)∧(1/x∧3)=limx→0e∧ln（sinx/x)^(1/x^2)=e∧limx→0(lnsinx-lnx)/x^2(这是0/0型,运用洛必达法则）=e∧limx→0(cosx/sinx-1/x)/2x=e∧

如何求limx→0（3^x-1)/xlim(x→0)(3^x-1)/x=lim(x→0)(3^x-1)'/x'=lim(x→0)(3^xln3)=ln3洛必达法则，原式x趋于0，=xln3/1=0limx->0(3^x*ln3)/1=ln3

limx→∞(2^x-1)/3^x=? 

limx→∞(1+2/x)^(x+3)求极限 lim(x→∞)(1+2/x)^(x+3)lim(x→∞)(1+2/x)^[(x/2)*2(x+3)/x]=lim(x→∞)[(1+2/x)^(x/2)]^(2(x+3)/x)=e^(

limx→∞(2x^3-x+1)limx->02x^3=0limx->0x=0limx->01=1把三个加起来=1

limx→0(1-x)^(1/x)是1/e.原式=e^(ln（1-x)/x)=e^(-1)=1/e

limx→0{(tanx-x)/x^3}lim(x→0){(tanx-x)/x^3}=lim(x→0{(tanx-x)'/(x^3)'=lim(x→0{(1/cosx^2-1)'/(3x^2)'=lim(x→0)(2sinx/cosx^3)

limx→0(1-x)^x=x→0时分子趋向于1,分母趋向于0.1^0=1.=11111111111111111111111111111111

求极限1.limx→-1(x^3+1)/sin(x+1);2.limx→0(e^x-e^-x)/(sinx);3.limx→+∞(ln(1+1/x))/(arccotx);4.limx→∞(tanx-sinx)/(1-cos2x);1、li

limx→1x^2-x+1/x-1=limx→0+x^sinx=limx→1(x^2-x+1)/(x-1)分子极限为1,分母极限为0,∴limx→1(x^2-x+1)/(x-1)=∞limx→0+x^sinx设y=x^sinx,取对数得,l

求极限limx→∞(x+1)(x+2)(x+3)(x+4)/x^4limx→∞(x+1)(x+2)(x+3)(x+4)/x^4=limx→∞(1+1/x)(1+2/x)(1+3/x(1+4/x)=1

limx→0(1+3x)^(1/sinx)求解答.