(cosx)^(1/x^2)求当x趋向于0时的极限 要过程 求极限 x 趋于0 lim(cosx)^1/(x^2) 求步...
(cosx)^(1/x^2)\u6c42\u5f53x\u8d8b\u5411\u4e8e0\u65f6\u7684\u6781\u9650lim ln(cosx)^(1/x^2)
=lim(ln(cosx)/x²)
=lim(-sinx/cosx)/(2x)
=-lim(sinx/(2xcosx)
=-1/2
\u4e0d\u53ef\u4ee5\u76f4\u63a5\u628ax=0\u4ee3\u5165\uff0c\u56e0\u4e3ax\u662f\u8d8b\u5411\u4e8e0\u800c\u4e0d\u662f\u7b49\u4e8e0\uff0c\u4e5f\u4e0d\u5141\u8bb8\u51fa\u73b0\u88ab0\u9664\u7684\u60c5\u51b5\u3002
\u4e25\u683c\u6765\u8bf4\uff0c\u4efb\u4f55\u65f6\u5019\u90fd\u4e0d\u80fd\u628a0\u4ee3\u5165\uff0c\u90fd\u662f\u8fd0\u7528\u6781\u9650\u503c\u6765\u8fd0\u7b97\u3002
\u5229\u7528\u5bf9\u6570\u6027\u8d28
(cosx)^(1/x^2)=e^[ln(cosx)^(1/x^2)]
=e^(1/x^2 * lncosx)
=e^(lncosx/x^2)
\u53ea\u8981\u5bf9\u6307\u6570\u90e8\u5206\u6c42\u6781\u9650\u5373\u53ef\uff0c\u6709\u4e24\u79cd\u65b9\u6cd5\uff1a
\u4e00\u3001\u7b49\u4ef7\u65e0\u7a77\u5c0fln(1+x)\uff5ex,1-cosx\uff5e x^2/2
lim(lncosx/x^2)=lim ln[1+(cosx-1)]/x^2
=lim (cosx-1)/x^2
=lim (-x^2/2)/x^2
=-1/2
\u4e8c\u3001\u5229\u7528\u6d1b\u5fc5\u8fbe\u6cd5\u5219\u5206\u5b50\u5206\u6bcd\u6c42\u5bfc\u53ca\u516c\u5f0flim sinx/x=1
lim(lncosx/x^2)=lim (-sinx/cosx)/2x
=lim (-1/2cosx)
=-1/2
\u6240\u4ee5\u539f\u5f0f=lim e^(lncosx/x^2)
=e^lim(lncosx/x^2)
=e^(-1/2)
\u6269\u5c55\u8d44\u6599\uff1a
\u4e00\u4e2a\u51fd\u6570\uff0c\u53ef\u4ee5\u5b58\u5728\u4e0d\u5b9a\u79ef\u5206\uff0c\u800c\u4e0d\u5b58\u5728\u5b9a\u79ef\u5206\uff0c\u4e5f\u53ef\u4ee5\u5b58\u5728\u5b9a\u79ef\u5206\uff0c\u800c\u6ca1\u6709\u4e0d\u5b9a\u79ef\u5206\u3002\u8fde\u7eed\u51fd\u6570\uff0c\u4e00\u5b9a\u5b58\u5728\u5b9a\u79ef\u5206\u548c\u4e0d\u5b9a\u79ef\u5206\u3002
\u82e5\u5728\u6709\u9650\u533a\u95f4[a\uff0cb]\u4e0a\u53ea\u6709\u6709\u9650\u4e2a\u95f4\u65ad\u70b9\u4e14\u51fd\u6570\u6709\u754c\uff0c\u5219\u5b9a\u79ef\u5206\u5b58\u5728\uff1b\u82e5\u6709\u8df3\u8dc3\u3001\u53ef\u53bb\u3001\u65e0\u7a77\u95f4\u65ad\u70b9\uff0c\u5219\u539f\u51fd\u6570\u4e00\u5b9a\u4e0d\u5b58\u5728\uff0c\u5373\u4e0d\u5b9a\u79ef\u5206\u4e00\u5b9a\u4e0d\u5b58\u5728\u3002
\u6c42\u51fd\u6570f(x)\u7684\u4e0d\u5b9a\u79ef\u5206\uff0c\u5c31\u662f\u8981\u6c42\u51faf(x)\u7684\u6240\u6709\u7684\u539f\u51fd\u6570\uff0c\u7531\u539f\u51fd\u6570\u7684\u6027\u8d28\u53ef\u77e5\uff0c\u53ea\u8981\u6c42\u51fa\u51fd\u6570f(x)\u7684\u4e00\u4e2a\u539f\u51fd\u6570\uff0c\u518d\u52a0\u4e0a\u4efb\u610f\u7684\u5e38\u6570C\u5c31\u5f97\u5230\u51fd\u6570f(x)\u7684\u4e0d\u5b9a\u79ef\u5206\u3002
\u5c06\u6240\u6c42\u79ef\u5206\u5316\u4e3a\u4e24\u4e2a\u79ef\u5206\u4e4b\u5dee\uff0c\u79ef\u5206\u5bb9\u6613\u8005\u5148\u79ef\u5206\u3002\u5b9e\u9645\u4e0a\u662f\u4e24\u6b21\u79ef\u5206\u3002
\u6709\u7406\u51fd\u6570\u5206\u4e3a\u6574\u5f0f\uff08\u5373\u591a\u9879\u5f0f\uff09\u548c\u5206\u5f0f\uff08\u5373\u4e24\u4e2a\u591a\u9879\u5f0f\u7684\u5546\uff09\uff0c\u5206\u5f0f\u5206\u4e3a\u771f\u5206\u5f0f\u548c\u5047\u5206\u5f0f\uff0c\u800c\u5047\u5206\u5f0f\u7ecf\u8fc7\u591a\u9879\u5f0f\u9664\u6cd5\u53ef\u4ee5\u8f6c\u5316\u6210\u4e00\u4e2a\u6574\u5f0f\u548c\u4e00\u4e2a\u771f\u5206\u5f0f\u7684\u548c\uff0e\u53ef\u89c1\u95ee\u9898\u8f6c\u5316\u4e3a\u8ba1\u7b97\u771f\u5206\u5f0f\u7684\u79ef\u5206\uff0e
\u53ef\u4ee5\u8bc1\u660e\uff0c\u4efb\u4f55\u771f\u5206\u5f0f\u603b\u80fd\u5206\u89e3\u4e3a\u90e8\u5206\u5206\u5f0f\u4e4b\u548c\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1--\u4e0d\u5b9a\u79ef\u5206
根据题意回答：
x→0
原极限=e^lim ln (cosx)^(1/x^2)
lim ln(cosx)^(1/x^2)
=lim ln(1+cosx-1) / x^2
利用等价无穷小：
ln(1+x)~x，1-cos~x^2/2
=lim (-x^2/2)/x^2
=-1/2
极限单调性：
单调有界准则：单调增加（减少）有上（下）界的数列必定收敛。
在运用以上两条去求函数的极限时尤需注意以下关键之点。
一是先要用单调有界定理证明收敛，然后再求极限值。
二是应用夹挤定理的关键是找到极限值相同的函数，并且要满足极限是趋于同一方向 ，从而证明或求得函数的极限值。
x→0
原极限=e^lim ln (cosx)^(1/x^2)
考虑lim ln(cosx)^(1/x^2)
=lim ln(1+cosx-1) / x^2
利用等价无穷小：ln(1+x)~x，1-cos~x^2/2
=lim (-x^2/2)/x^2
=-1/2
原极限=e^(-1/2)有不懂欢迎追问
这是1的无穷强大次幂 型
原式=e^((cosx-1)/x²) ，x趋向于0
(cosx-1)/x²罗比达法则-sinx/(2x)=-1/2
原式=e^(-1/2)
x→0
原极限=e^lim ln (cosx)^(1/x^2）
因为lim ln(cosx)^(1/x^2) =lim ln(1 cosx-1) / x^2 利用等价无穷小：ln(1 x)~x，1-cos~x^2/2 =lim (-x^2/2)/x^2 =-1/2 原极限=e^(-1/2)
绛旓細=lim(x鈫0)(1+cosx-1)^锛1+cot^2x锛=lim(x鈫0)(1+cosx-1)^[1/(cosx-1)][(cosx-1)(1+cot^2x)]搴曟暟锛歭im(x鈫0)(1+cosx-1)^[1/(cosx-1)]=e 鎸囨暟锛歭im(x鈫0)[(cosx-1)(1+cot^2x)]=lim(x鈫0)[(cosx-1)(sin^2x+cos^2x)]/sin^2x =lim(x鈫0)(cosx-1)/sin^...
绛旓細f(x)=(cosx)^(1/(ln(1+x^2));浠ヤ笅绛夊彿鐪佺暐鏋侀檺绗﹀彿 lnf(x)=(1/(ln(1+x^2))lncosx=[1/x^2]*ln(cosx-1+1)=1/x^2*[-x^2/2]=-1/2;鍒╃敤浜嗙瓑浠烽噺.ln锛1+x锛墌x;cosx-1~-x^2/2 f(x)=e^(-1/2)
绛旓細涓銆佺瓑浠锋棤绌峰皬ln(1+x)锝瀤,1-cosx锝 x^2/2 lim(lncosx/x^2)=lim ln[1+(cosx-1)]/x^2 =lim (cosx-1)/x^2 =lim (-x^2/2)/x^2 =-1/2 浜屻佸埄鐢ㄦ礇蹇呰揪娉曞垯鍒嗗瓙鍒嗘瘝姹傚鍙婂叕寮弆im sinx/x=1 lim(lncosx/x^2)=lim (-sinx/cosx)/2x =lim (-1/2cosx)=-1/2 鎵浠ュ師寮...
绛旓細娉瑧涓烘偍瑙ｇ瓟锛屽鑻ユ弧鎰,璇风偣鍑籟閲囩撼涓烘弧鎰忓洖绛擼;濡傝嫢鎮ㄦ湁涓嶆弧鎰忎箣澶,璇锋寚鍑,鎴戜竴瀹氭敼姝!甯屾湜杩樻偍涓涓纭瓟澶!绁濇偍瀛︿笟杩涙!
绛旓細绠鍗曡绠椾竴涓嬪嵆鍙紝绛旀濡傚浘鎵绀
绛旓細鍘熷紡=exp{lim(x->0)ln(cosx^2)/sinx^2}=exp{lim(x->0)ln[1+(cosx^2-1)]/sinx^2}=exp{lim(x->0)(cosx^2-1)/x^2}=exp{lim(x->0)-(x^4/2)/x^2}=exp(0)=1 杩欓噷鐢ㄤ簡绛変环鏃犵┓灏弆n(1+x)~x,1-cosx~x^2/2 (x->0)濡傛灉棰樹腑涓篶os^2x涓巗in^2x,鍒欐湁鏇寸畝鍗曠殑鏂规硶...
绛旓細lim ln(cosx)^(1/x^2)=lim ln(1+cosx-1) / x^2 鍒╃敤绛変环鏃犵┓灏忥細ln(1+x)~x锛1-cos~x^2/2 =lim (-x^2/2)/x^2 =-1/2 鏋侀檺鍗曡皟鎬э細鍗曡皟鏈夌晫鍑嗗垯锛氬崟璋冨鍔狅紙鍑忓皯锛夋湁涓婏紙涓嬶級鐣岀殑鏁板垪蹇呭畾鏀舵暃銆傚湪杩愮敤浠ヤ笂涓ゆ潯鍘绘眰鍑芥暟鐨勬瀬闄愭椂灏ら渶娉ㄦ剰浠ヤ笅鍏抽敭涔嬬偣銆備竴鏄厛瑕佺敤鍗曡皟鏈夌晫瀹氱悊...
绛旓細鍘熸瀬闄=e^limx鈫0 ln (cosx)^(1/x^2)鑰冭檻lim ln(cosx)^(1/x^2)=lim ln(1+cosx-1) / x^2 鍒╃敤绛変环鏃犵┓灏忥細ln(1+x)~x,1-cos~x^2/2 =lim (-x^2/2)/x^2 =-1/2
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細
绛旓細lim(x->0)锛坈osx锛塣锛 1/x锛=lim(x->0)[1+(cosx-1)]^锛 1/x锛=e^lim(x->0)(cosx-1)/x =e^lim(x->0)(-x^2/2)/x =e^lim(x->0)(-x/2)=e^0 =1
