limx→01x
limx→∞arctanx/xx→∞时,arctanx是有界量,1/x是无穷小量,所以limx→∞arctanx/x=0
limx→osin5x/x原式=5*lim(x->0)sin5x/5x=5*1=5
limx→osin5x/x原式=5*lim(x->0)sin5x/5x=5*1=5
limx→0(1-x)^x=x→0时分子趋向于1,分母趋向于0.1^0=1.=11111111111111111111111111111111
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→-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→0(arctanx/x)极限步骤用罗必达法则,一次就出来了.
limx→0+(x^sinx)求极限limx→+0时,tan9x等价于9x,sin√x等价于√x,sinx^2等价于x^2原式=(9x)^3/2*√x/(x^2)=27这样可以么?
limx→0(tanx-sinx)/xlim(x→0)(tanx-sinx)/x=lim(x→0)tanx(1-cosx)/x=lim(x→0)(1-cosx)=0洛比达法则分子分母同时求导原式=limx→0(1/(cosx)^2-cosx
limx→0x/根号(1-cosx)lim(x→0)x/√(1-cosx)=lim(x→0)x/√2sin²x/2=lim(x→0)2/√2*(x/2)/(sinx/2)=2/√2=√2【数学辅导团】为您解答√2,1-cosx~1
limx→0(x/sin2x)=
limx→1lnx/x-1=limlnx/(x-1)=lim(lnx)'/(x-1)'=lim(1/x)/1=1limx趋近于1,所以x趋近于1,lnx趋近于0,0/x等于0,0-1等于-1如果是(x-1)答案为1如果是整体减1答案为0li
limx→0(cotx-x分之一)通分:lim(cotx-1/x)=lim(xcosx-sinx)/(xsinx)由等价无穷小代换,sinx∴原式=lim(xcosx-sinx)/x²0/0型,∴用罗必塔法则:=lim(cosx-
limx→0(a^-1)/xlimx→0(a^x-1)/x=lna
limx→0tan3x/x计算极限x趋于0,tanx和x是等价无穷小则tan3x和3x是等价无穷小tan3x/x=3*(tan3x/3x)所以极限=3
limx→1arcsin(1-x)/lnx
limx→0(a^x-1)/xlna洛必达定理,分子求导成lna*a^X,分子变成lna,相除变成a^X,当X=0时,极限为1
limx→0(1/x)^tanx原式=e^{lim(x->0)[ln(1/x)/cotx]}=e^{lim(x->0)[(x(-1/x²))/(-csc²x)]}(∞/∞性极限,应用罗比达法则)=e^{lim(x->0)
limx→0xcos1/x等于多少无穷小乘有界函数,还是无穷小所以结果为00cos1/x有界