lim(x→∞)(cos1/x)^x
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/18 10:30:31
lim(x→∞)(cos1/x)^xlim(x→∞)(cos1/x)^xlim(x→∞)(cos1/x)^x令y=(cos1/x)^xlny=xlncos(1/x)则limlny=limlncos(1
lim(x→∞)(cos1/x)^x
lim(x→∞)(cos1/x)^x
lim(x→∞)(cos1/x)^x
令y=(cos1/x)^x
lny=xlncos(1/x)
则limlny
=limlncos(1/x)/(1/x)
是∞/∞型,用洛必达法则
=lim[1/cos(1/x)*(-sin1/x)*(1/x)']/(1/x)'
=lim[-tan(1/x)]
=0
所以原式=e^0=1
lim(x→∞)(cos1/x)^x
X→0,lim(sin1/x+cos1/x)^x
求极限,请高手指导.lim(x→+∞) [(sin1/x + cos1/x)]^x
求极限lim(x→∞)(sin1/x-cos1/x)^x
求极限lim(x→∞)(sin1/x-cos1/x)^x
lim(cos1/n)^n^2 n->∞lim (1+|x|)^1/x x->0
求lim x->无穷(sin2/x+cos1/x)^x
lim [2x(sin1/x)-(cos1/x)]/cosx 为什么极限不存在?x→0
lim(cos1/√x)^2x(x->无穷)
lim(x趋向无穷）(cos1/根号x)^2x
求极限lim(x→+0)(lnx+cos1/x)/(lnx+sin1/x) 请不要用洛必达法则
求极限lim(x→+0)(lnx+cos1/x)/(lnx+sin1/x) 请不要用洛必达法则
lim(x趋近于0)cos1/x为多少
lim x→0 sin2x/sin5xlim x→0 (1-cos2x)/x²lim x→0 in(1-x/2)^1/xlim x→∞ (1+2/x)^2x+1lim x→0 sin2x/sin5xlim x→0 (1-cos2x)/x²lim x→0 in(1-x/2)^1/xlim x→∞ (1+2/x)^2x+1lim x→3 (x-3)cos1/(x-3)lim x→0 (e^sinx-e^x)/(sinx-x)
lim(cos1/x)^x^2limX趋于无穷(cos1/x)^x^2 x趋于无穷,结果是答案是e^-0.5
利用无穷小的性质,计算下列极限(1)limx平方cos1/x 下面是x→0(2)lim arctanx/x 下面是x→∞
lim(x→无穷)上面(3sinx+x^2cos1/x)下面(1+cosx)ln(x-1)
求lim[1+(sin1/x+cos1/x-1)]^x (x趋于无穷大)