求lim的极限
L =lim(x->1) (2-x) ^[sec(πx\\\/2)]lnL =lim(x->1) ln(2-x) \\\/ [cos(πx\\\/2)] (0\\\/0)=lim(x->1) [-1\\\/(2-x)] \\\/ [ -(π\\\/2) sin(πx\\\/2)] = 2\\\/πL = e^(2\\\/π)
求极限 lim
(1)当y→2时，分式为0\\\/0用洛必则对上下分别求导：原式=lim y→2 3(y-2)^2\\\/4y^3=0(2)请确定题目是否错误，当x→3时内的式子为负数，无法开根号。
若x→-3，原式=lim x→-3 [(x+2)(x+3)\\\/(x-4)(x+3)]^1\\\/4 =lim x→-3[(x+2)\\\/(x-4)]^1\\\/4 =(1\\\/7)^1\\\/4(3)分子有理化，原式=lim x→0 [(x+4) -4]\\\/x([√(x+4) +2] =lim x→0 1\\\/[√(x+4) +2] =1\\\/4(4)分母有理化，原式=lim x→0 x[(√(x+2) +√2]\\\/x =lim x→0 (√(x+2) +√2] =2√2
求极限lim。
x->0cosx ~ 1- (1\\\/2)x^2x-xcosx ~ (1\\\/2)x^3sinx~ x -(1\\\/6)x^3x-sinx ~ (1\\\/6)x^3lim(x->0) ( x-xcosx)\\\/(x-sinx)=lim(x->0) (1\\\/2)x^3 \\\/[ (1\\\/6)x^3]=3
