lim[cos(a/x)]^(x^2) x趋近于无穷大时求极限值

来源：学生作业帮助网编辑：作业帮时间：2024/11/06 05:23:41

lim[cos(a/x)]^(x^2) x趋近于无穷大时求极限值
lim[cos(a/x)]^(x^2) x趋近于无穷大时求极限值

lim[cos(a/x)]^(x^2) x趋近于无穷大时求极限值
lim[cos(a/x)]^(x^2)
=lim e^[x^2*lncos(a/x)]
=e^{lim x^2[cos(a/x)-1]}
=e^{lim x^2[-(a/x)^2/2]}
=e^[lim -a^2/2]
=e^(-a^2/2)

f(x)=[cos(a/x)]^(x^2)
lnf(X)=x^2ln[cos(a/x)]
令t=1/x
g(t)=lnf(x)=ln[cos(at)]/t^2 t趋向于0时
g(t)是0/0型
用罗必塔法则有
lim g(t)=-sin(at)*a/(2t*cos(at))=-a^2cosat/2=-a^2/2 ,t趋向于0
所以limf(x)= e^limg(t)=e^(-a^2/2)

lim( cos(a+2x)-2cos(a+x)+cosa)/x^2x->0 求极限 lim[cos(a/x)]^(x^2) x趋近于无穷大时求极限值 lim[cos(a/x)]^(x^2) x趋近于无穷大时求极限值 lim(x→0)(cosx-cos^2x)/x^2 lim(x->PI/2) COS X/(PI/2)-X lim(sin(x^2*cos(1/x)))/x怎么做? lim x趋向0 (1+cos^2(1/x))^x 极限题,1.lim x-->0 cos(2x)-cos(3x)/x^22.lim x-->∞ cos(2x)-cos(3x)/x^2 如何求 lim(n→∞)cos(x/2)cos(x/4)...cos(x/2n) lim x->0 cos(2x)-1 / cosx -1 = lim[cos(1/x)]^2,x--->0,极限为什么不存在求lim(x→0)x cos 1/x lim(x→∞)x^2/ (3x-1)的极限 lim(cos(sqrt(x)))^1/x 当x→0时,下列函数哪一个是其他三个的高阶的无穷小A x² B 1-cosx C x-tanx D In(1+x²)lim (x-tanx)/x²=lim (1-/cos²x)/2x=lim ((cos²x-1)/2x).(1/cos²x)=-lim (sin²x/2x).1/cos²x=0倒数第二步是 lim x->0 1-cos(3x)/(2x) 和 lim x->0 1-cos3x/2x^2 求lim (sin^2x+x)/(cos^2x-x) x趋向于无穷 lim(n→∞) (cos x/n)^n^2 求lim (sec^2 x求极限lim (sec^2 x - 1 )/(1-cos x) X-0请写出过程,