lim 1-cos(x)/sin(5x) x是趋向0的.分子是1-cos(x) 分母是sin(5x)过程写一下,

lim(sin(x^2*cos(1/x)))/x怎么做? 用罗必达法则解 （见图）lim [(e^x)-1/sin（5x)]x->0lim [1-cos(5x)/1-cos(7x)]x->0lim x->0 [(8^x-7^x-1)/(x^2)-1]lim [1-(6/x)]^x x->∞ lim 7cos(3x)sec(9x)x->pi/2 lim 1-cos(x)/sin(5x) x是趋向0的.分子是1-cos(x) 分母是sin(5x)过程写一下, (x^2cos^2x-sin^2x)/x^2cos^2xsin^2x （x->0）lim (x^2cos^2x-sin^2x)/x^2cos^2xsin^2x （x->0）=lim [cos^2x-(sin^2x/x^2)]/cos^2xsin^2x=lim (cos^2x-1)/sin^2x=lim -sin^2x/sin^2x=-1这样解错在哪啊?小弟感激不尽 求 lim(x→∞)[sin(2/x)+cos(1/x)]^x的极限. lim(x→0) [1-cos(x^2)]/sin(x^2)tan(x^2)]=? lim(sin x+cos x)^（1／X）(X趋近于零） lim(x趋近于0)(（1/sin^2 x）-(cos^2 x)/x^2)= lim(x趋向于0）(1+sinx-cosx)/(1+sinβx-cosβx) lim(Inx+cos(1/x))/(Inx+sin(1/x))x趋近于0+. X趋向于π,求lim((sin^2 X)/()1+cos^3 X) lim(x→0)(cos x)∧(1/sin 2x)＝? 【【【高数】】】这道极限怎么解?lim(sin(2/x)+cos(1/x))^x 用洛必达法则求几个极限（有图）~用洛必达法则求以下极限（见图）lim [(e^x)-1/sin（5x)]x->0lim [1-cos(5x)/1-cos(7x)]x->0lim x->0 [(8^x-7^x-1)/(x^2)-1]lim [1-(6/x)]^x x->0 lim 7cos(3x)sec(9x)x->∞（搞错了~应该是>/ cos x/sin x+sin x/(1+cos x) 化简. lim(cos(sqrt(x)))^1/x 当 x->0 ,且 x>0 时,求极限：lim ln(2*cos(x)+sin(1/x)) . lim(x(sin)^2)