limx→0（1-cos2x）/x(（e^3x）-1)求极限呢？ limx→0（(e^x)-1）/2x求极限呢？

limx\u21920\uff081-cos2x\uff09/x(\uff08e^3x\uff09-1)\u6c42\u6781\u9650\u5462\uff1f

\u6b64\u5904\u53ef\u4ee5\u5229\u7528\u7b49\u4ef7\u65e0\u7a77\u5c0f\u7684\u4ee3\u6362\u6765\u8ba1\u7b97\uff0c\u6ce8\u610f\u4f7f\u7528\u65f6\u5fc5\u987b\u6ee1\u8db3\u4f60\u66ff\u6362\u7684\u90a3\u90e8\u5206\u4e0e\u5269\u4f59\u90e8\u5206\u662f\u4e58\u6216\u8005\u9664\u7684\u5173\u7cfb\uff0c\u5177\u4f53\u8fc7\u7a0b\u770b\u56fe\uff1a

\u8fd9\u662f\u5c5e\u4e8e\u76f4\u63a5\u7528\u7b49\u4ef7\u65e0\u7a77\u5c0f\u66ff\u6362\u6c42\u6781\u9650\u7684\u5e38\u89c4\u6027\u95ee\u9898\u3002

求极限，当x→0，（1-cos2x）/x(e^3x-1)
方法1：使用等价无穷小
e^x=1+x，1-cos2x=0.5*4x^2
原式=2x^2/x/3x=2/3

这仍然是无穷小的等价替换
在求极限中的应用。

lim(x鈫0)(1-cos2x)/(xsinx)鐨勬瀬闄
绛旓細cos2x=2cos^x-1=1-2sin^x 閭ｄ箞鍒嗗瓙灏辩瓑浜2sin^x锛屽張鍥犱负x鈫0锛宻inx/x=1,鎵浠inx=x 甯﹀叆 缁撴灉涓2

x鈫0鏃,1-cos2x绛変环浜1/2(2x)²=2x²? 鎶1/2涔樿繘鍘讳笉鏄痻^2鍚...
绛旓細=2sin²x/xsinx =2sinx/x lim(x鈫0) (1-cos2x)/xsinx=lim(x鈫0) 2sinx/x=2 鍏堝埄鐢 x绛変环浜巗inx鏈 鍘熷紡 = lim (1-cos2x)/x²=lim 2sin2x / 2x (娲涙瘯濉旀硶鍒)=lim 2x/x =2 lim(x鈫0) (1-cos2x)/xsinx =lim(x鈫0) [2(sinx)^2]/xsinx =lim(x鈫0) 2...

lim(x鈫0)(1-cos2x)/(x*sin2x) lim(x鈫0)(xcot2x) lim(x鈫掓棤绌)(1+...
绛旓細lim(x鈫0)(1-cos2x)/(x*sin2x)=lim(x鈫0)2sin²x/(x*sin2x)鐢ㄧ瓑浠锋棤绌峰皬 =lim(x鈫0)2x²/(x*2x)=1 lim(x鈫0)(xcot2x)=lim(x鈫0)x/tan2x 鐢ㄧ瓑浠锋棤绌峰皬 =lim(x鈫0)x/2x =1/2 lim(x鈫掆垶)(1+(1/2x))^x =lim(x鈫掆垶)(1+(1/2x))^(2x/2)鐢ㄩ噸瑕...

lim(x鈫0)[(1-cos2x)/(xsinax)]
绛旓細1-cos2x=2sin^2(x)绛変环浜庢棤绌峰皬2x^2 sinx绛変环浜巟 鍘熷紡=2/1=2

limx瓒嬭繎浜0銆1-cos(2x)銆/(xsinx)
绛旓細鍥犱负cos2x=cos²x-sin²x=1-2sin²x锛屽垯鏈1-cos2x=2sin²x 鎵浠(1-cos2x)/(xsinx)=(2sin²x)/(xsinx)=2sinx/x锛屾牴鎹袱涓噸瑕佹瀬闄愶紝寰楀嚭缁撴灉=2

鏋侀檺lim x鈫o(1-cos2x)/xtanx=
绛旓細绛変环鏃犵┓灏忕殑鏇挎崲锛岀粨鏋滀负2銆

姹lim(x鈫0)銆1-cos2x-(tan^2) X銆/xsinx 绛旀鏄3 姹傝繃绋
绛旓細鍦▁瓒嬩簬0鐨勬椂鍊欙紝1-cosx绛変环浜0.5x^2 鎵浠ュ湪杩欓噷锛1-cos2x绛変环浜0.5(2x)^2鍗2x^2 鑰 tanx鍜宻inx閮界瓑浠蜂簬x锛屾墍浠ュ緱鍒板師鏋侀檺 =lim(x鈫0)(2x^2 -x^2)/x^2 = 1 绛旀搴旇鏄1鐨勫惂锛熶細涓嶄細浣犵鍙峰啓閿欎簡锛

璁lim x鈫0 ,[x路f(x)/1-cos2x]=1,鍏朵腑f(0)=0,鍒檉'(0)=?
绛旓細褰搙鈫0鏃讹紝1 - cos(2x) 鈫 0 瑕佷护杩欐瀬闄愭湁鎰忎箟锛屽垯鍒嗗瓙浜﹁ 鈫 0锛屽嵆xƒ(x) 鈫 0 lim(x鈫0) xƒ(x)/(1 - cos(2x))锛<-- 0/0褰㈠紡锛屾礇蹇呰揪娉曞垯 = lim(x鈫0) [ƒ(x) + xƒ'(x)]/(2sin(2x)) <-- 0/0褰㈠紡锛屽啀娲涘繀杈炬硶鍒 = lim(x鈫0) ...

璁lim x鈫0 ,[x路f(x)/1-cos2x]=1,鍏朵腑f(0)=0,鍒檉'(0)=?
绛旓細璁lim(x鈫0)xƒ(x)/(1 - cos(2x))= 1锛屽叾涓ƒ(0)= 0锛屽垯ƒ'(0)= 锛熷綋x鈫0鏃讹紝1 - cos(2x)鈫 0 瑕佷护杩欐瀬闄愭湁鎰忎箟锛屽垯鍒嗗瓙浜﹁ 鈫 0锛屽嵆xƒ(x)鈫 0 lim(x鈫0)xƒ(x)/(1 - cos(2x))锛<-- 0/0褰㈠紡锛屾礇蹇呰揪娉曞垯 = lim(x鈫0)[ƒ...

姹lim(x鈫0)銆1-cos2x+tan^2 X銆/xsinx
绛旓細鍏堣浆鍖1-cos2x=2sin^2x 鍖栦负涓ら儴鍒2sin^2x/xsinx鍜宼an^2x/xsinx 鍓嶉潰涓閮ㄥ垎绾﹀幓sinx鍚庣敤甯歌鏋侀檺sinx/x=1,鍓嶅崐閮ㄥ垎涓2 鍚庡崐閮ㄥ垎鏄(tanx/x)(1/cosx),鍓嶅悗閮芥槸甯歌鏋侀檺绛変簬1锛屽悗鍗婇儴鍒嗙瓑浜1 鏈鍚庣殑缁撴灉灏辨槸2+1=3

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