求极限 lim (coshx+cosx-2)/{[(sinhx)^2][(sinx)^2)]},x趋近于0
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/02 17:46:07
求极限 lim (coshx+cosx-2)/{[(sinhx)^2][(sinx)^2)]},x趋近于0
∵lim(x->0)[(coshx+cosx-2)/x^4]
=lim(x->0)[(sinhx-sinx)/(4x^3)] (0/0型极限,应用罗比达法则)
=lim(x->0)[(coshx-cosx)/(12x^2)] (0/0型极限,应用罗比达法则)
=lim(x->0)[(sinhx+sinx)/(24x)] (0/0型极限,应用罗比达法则)
=lim(x->0)[(sinhx/x+sinx/x)/24]
=(1+1)/24 (应用重要极限lim(x->0)(sin/x)=1,lim(x->0)(sinh/x)=1)
=1/12
∴lim(x->0)[(coshx+cosx-2)/(((sinhx)^2)*((sinx)^2))]
=lim(x->0){[(coshx+cosx-2)/x^4]*[(x/sinhx)^2]*[(x/sinx)^2]}
={lim(x->0)[(coshx+cosx-2)/x^4]}*{[lim(x->0)(x/sinhx)]^2]}*{[lim(x->0)[(x/sinx)]^2}
=(1/12)*(1^2)*(1^2) (应用重要极限lim(x->0)(sin/x)=1,lim(x->0)(sinh/x)=1)
=1/12.
=lim(x->0)[(sinhx-sinx)/(4x^3)] (0/0型极限,应用罗比达法则)
=lim(x->0)[(coshx-cosx)/(12x^2)] (0/0型极限,应用罗比达法则)
=lim(x->0)[(sinhx+sinx)/(24x)] (0/0型极限,应用罗比达法则)
=lim(x->0)[(sinhx/x+sinx/x)/24]
=(1+1)/24 (应用重要极限lim(x->0)(sin/x)=1,lim(x->0)(sinh/x)=1)
=1/12
∴lim(x->0)[(coshx+cosx-2)/(((sinhx)^2)*((sinx)^2))]
=lim(x->0){[(coshx+cosx-2)/x^4]*[(x/sinhx)^2]*[(x/sinx)^2]}
={lim(x->0)[(coshx+cosx-2)/x^4]}*{[lim(x->0)(x/sinhx)]^2]}*{[lim(x->0)[(x/sinx)]^2}
=(1/12)*(1^2)*(1^2) (应用重要极限lim(x->0)(sin/x)=1,lim(x->0)(sinh/x)=1)
=1/12.
求极限 lim (coshx+cosx-2)/{[(sinhx)^2][(sinx)^2)]},x趋近于0
求极限:lim(sinx)^tanx (x趋近于pai/2)
求极限 lim e^x^2 - 1 / cosx - 1 其中x趋近于0
求lim x趋近于0 sinx^3 tanx/(1-cosx^2)
lim趋近于0((3+2sinx)*x-3*x)/((tanx)*2)求极限
lim趋近于0((3+2sinx)*x)/((tanx)*2)求极限
计算极限 lim(sin3x/sin5x) x趋近于0 lim[2x²/(1-cosx)]x趋近于0
求Lim(sinx+cosx)1/x次方x趋近于0的极限
lim(sinx-xcosx)/x(1-cosx)用洛必达法则求极限(x)趋近于0
lim {(cosx)^(1/2)-(cosx)^(1/3)}/ (sinx)^2 x趋近于0
求LIM(1-COSX)/X*SINX X趋近于0
求极限,当x趋近于0时,lim{(e^(2x)-e^(-x)-3x)/(1-cosx)}的值