lim1-cosx

晏趴饰4139lim1 - cosx/xtanx=求过程 -
扈肩尹13127243972 ______ lim(X->0)[(1-cosX)/(XtanX)]=lim(X->0)[sin(x/2)]^2/(XtanX)=lim(X->0)(X/2)^2/(X*X)=1/4

晏趴饰4139当x趋于0时lim(1 - cosx)/(1+x*x)的极限怎么求 -
扈肩尹13127243972 ______[答案] 由于把X=0直接带入分母不为零, 则分母趋近于1+0=0 分子由于lim(X->0)cosX=1 所以分子趋近于1-1=0 各个数都有极限,且符合分母不为0 分子为0 分母为1 , 所以极限为0

晏趴饰4139lim[(1 - cosX)/x的平方]是多少,X趋近于0时 -
扈肩尹13127243972 ______[答案] 1- cosX = 1 - [ 1 - 2(sinx/2)^2] = 2[sin(x/2)]^2 1-cosX/X^2 = 2[sin(x/2)]^2/x^2 = (1/2)* [sin(x/2)/(x/2)]^2 sin(x/2)/(x/2) 当x趋近于0时,其极限为1 (这是书上的基本定理) 因此 lim[(1-cosX)/x^2] = 1/2

晏趴饰4139三角函数 极限lim (1 - cosx)/(2x^2)x - >0 -
扈肩尹13127243972 ______[答案] 洛必达法则: lim(1-cosx)/(2x^2) x->0 =lim(sinx/4x) x->0 =lim(cosx/4) x->0 =1/4

晏趴饰4139紧急!求lim(1 - cosx)^(2secx),x→π的极限 -
扈肩尹13127243972 ______[答案] 由于1-cosπ和2secπ均不等于0,所以不是不定式,可以直接代入π计算 原式=(1-cosπ)^(2secπ) =[1-(-1)]^[2(-1)] =2^(-2) =1/4

晏趴饰4139求极限lim(1 - cosx)/x^2 -
扈肩尹13127243972 ______[答案] x---0,1-cosx----0.5x^2; 所以答案就是0.5

晏趴饰4139求lim(1 - cosx)/x^2求lim(1 - COSX)/X^2X→0 -
扈肩尹13127243972 ______[答案] 答: lim(x→0) (1-cosx)/x² =lim(x→0) 2sin²(x/2)/[4*(x/2)²] =lim(t→0) (1/2)(sint/t)² =1/2

晏趴饰4139lim 1 - cosx/sinx计算题 -
扈肩尹13127243972 ______[答案] 方法一:三角函数法lim (1-cosx)/sinxx→0= lim [2sin²(½x)]/[2sin(½x)cos(½x)]x→0= lim [sin(½x)]/[cos(½x)]x→0= 0/1 = 0方法二:罗毕达求导lim (1-cosx)/sinxx→0= lim s...

晏趴饰4139limx趋向于0 (1 - cosx)/x² -
扈肩尹13127243972 ______[答案] 0/0型求导 limf/g=limf'/g' lim(1-cosx)/x2 =lim(1-cosx)'/(x^2 )' =lim(sinx)/2x =lim(sinx)'/(2x)' =lim(cosx/2) =1/2

晏趴饰4139为什么lim(1 - cosx)/[(3/2)x^2]会等于4/3*([sin(x/2)]^2)/x^2请给具体理由, -
扈肩尹13127243972 ______[答案] 由于二倍角公式:cosx=[cos(x/2)]^2-[sin(x/2)]^2······1 但由于[cos(x/2)]^2+[sin(x/2)]^2=1 所以1式有可以变成cosx=1-2[sin(x/2)]^2 所以 1-cosx=2[sin(x/2)]^2 代入上式lim(1-cosx)/[(3/2)x^2=lim{4/3*([sin(x/2)]^2)/x^2} 但不会直接=4/3*([sin(x/2)]^2)/x^2 ...