当x→0时,1-cos2x与什么为等价无穷小 已知当x→0时，1-cos2x与∫(sinx,0)ln(1+...
1-cos2x\u4e3a\u4ec0\u4e48\u7b49\u4ef7\u4e8e2x^2 \u5f53x\u8d8b\u4e8e\u96f6\u65f6\uff1f \u6709\u6ca1\u6709\u901a\u7528\u516c\u5f0f\uff1f\u901a\u7528\u516c\u5f0f\uff1a
cos2x\uff1d1\uff0d2sinx\uff3e2
cos2x\uff1d2cosx\uff3e2\uff0d1
1\uff0dcos2x\uff1dsin²x\uff0bcos²x²\uff0d\uff08cos²x\uff0dsin²x\uff09\uff1d2sin²x
sinx\u5728x\u8d8b\u5411\u4e8e0\u65f6\uff0c\u8d8b\u5411\u4e8ex
\u539f\u5f0f\uff1dlim2sin²x2x²
x\u21920\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff1d\uff0d\uff0d\uff0d\uff0d\uff0d\uff1d2
x²\uff0bxx²
cos2x\uff1dcos²x\uff0dsin²x\uff1d2cos²x\uff0d1\uff1d1\uff0d2sin²x
\u53ef\u4ee5\u6839\u636e\u60c5\u51b5\u9009\u62e9\u9700\u8981\u7684\u516c\u5f0f\u8fdb\u884c\u8ba1\u7b97\uff0e
\u6240\u4ee51\uff0dcos2x\uff1d1\uff0d\uff081\uff0d2sin²x\uff09\uff1d2sin²x\u3002
\u6269\u5c55\u8d44\u6599
\u4e3e\u4f8b\uff1a
\u8bbex\u8d8b\u5411\u4e8e0\u65f6\uff0c1\uff0dcos\uff3e2x\u4e0easin\uff3ex\uff0f2\u4e3a\u7b49\u4ef7\u65e0\u7a77\u5c0f\uff0c\u5219a
\u8bbex\u8d8b\u5411\u4e8e0\u65f6\uff0c1\uff0dcos\uff3e2x\u4e0easin\uff3ex\uff0f2\u4e3a\u7b49\u4ef7\u65e0\u7a77\u5c0f\uff0c\u5219a\uff1d\uff08\uff09
\u5df2\u77e5x\u21920\u65f6\uff0c1\uff0dcos\uff3e2x\u4e0easin\uff3e2\uff08x\uff0f2\uff09\u4e3a\u7b49\u4ef7\u65e0\u7a77\u5c0f
\u6240\u4ee5\uff1a
lim\uff081\uff0dcos\uff3e2x\uff09\uff0f\uff3basin\uff3e2\uff08x\uff0f2\uff09\uff3d\uff1d1
===> lim(sin^2 x)/[asin^2 (x/2)]=1
===> lim[2sin(x/2)*cos(x/2)]^2/[asin^2 (x/2)]=1
===> lim[4sin^2 (x/2)*cos^2 (x/2)]/[asin^2 (x/2)]=1
===> lim[4cos^2 (x/2)]/a=1
\u6240\u4ee5\uff0ca\uff1d4
1 - cos2x = 1 - (1 - 2sin²x) = 2sin²x\uff0c\u6700\u9ad8\u6b21\u6570\u662f2\uff0c\u6240\u4ee5\u5206\u5b50\u7684\u6700\u9ad8\u6b21\u6570\u4e5f\u662f2
lim(x\u21920) \u222b(0\u2192sinx) ln(1 + at) dt/(2sin²x)
= lim(x\u21920) cosxln(1 + asinx)/(4sinxcosx)
= lim(x\u21920) ln(1 + asinx)/(4sinx)
= lim(x\u21920) asinx/(4sinx)
= lim(x\u21920) a/4
\u7531\u4e8e\u662f\u7b49\u4ef7\u65e0\u7a77\u5c0f\uff0c\u4ee4\u6781\u9650\u7ed3\u679c\u7b49\u4e8e1
\u6240\u4ee5a = 4
lim(x→0)1-cos2x/∫(0→sinx)ln(1+at)dt
=lim(x→0)2x²/∫(0→sinx)ln(1+at)dt
=lim(x→0)4x/ln(1+asinx)*cosx
=lim(x→0)4x/asinx
=4/a
=1
所以
a=4
扩展资料
用极限思想解决问题的一般步骤可概括为：
对于被考察的未知量，先设法正确地构思一个与它的变化有关的另外一个变量，确认此变量通过无限变化过程的’影响‘趋势性结果就是非常精密的约等于所求的未知量；用极限原理就可以计算得到被考察的未知量的结果。
极限思想是微积分的基本思想，是数学分析中的一系列重要概念，如函数的连续性、导数（为0得到极大值）以及定积分等等都是借助于极限来定义的。如果要问：“数学分析是一门什么学科?”那么可以概括地说：“数学分析就是用极限思想来研究函数的一门学科，并且计算结果误差小到难于想像，因此可以忽略不计。
因为x→0时1-cosx等价于1/2 *x²
所以x→0时1-cos2x等价于1/2 *（2x）²
绛旓細x瓒嬩簬0鏃讹紝1-cos2x绛変环浜2x^2.
绛旓細lim(x鈫0)1-cos2x/鈭(0鈫sinx)ln(1+at)dt =lim(x鈫0)2x²/鈭(0鈫抯inx)ln(1+at)dt =lim(x鈫0)4x/ln(1+asinx)*cosx =lim(x鈫0)4x/asinx =4/a =1 鎵浠 a=4
绛旓細棣栧厛锛屾牴鎹嘲鍕掑睍寮鍏紡锛褰搙瓒嬭繎浜0鏃讹紝鎴戜滑鍙互灏哻os(2x)灞曞紑涓猴細cos(2x) 鈮 1 - (2x)^2/2! + (2x)^4/4! - (2x)^6/6! + ...灏嗗叾浠ｅ叆1 - cos(2x)鐨勮〃杈惧紡涓紝寰楀埌锛1 - cos(2x) 鈮 (2x)^2/2! - (2x)^4/4! + (2x)^6/6! - ...鎺ヤ笅鏉ワ紝鎴戜滑闇瑕佽绠梮ln(1+k...
绛旓細lim(x鈫0)1-cos2x/鈭(0鈫sinx)ln(1+at)dt =lim(x鈫0)2x²/鈭(0鈫抯inx)ln(1+at)dt =lim(x鈫0)4x/ln(1+asinx)*cosx =lim(x鈫0)4x/asinx =4/a =1 鎵浠 a=4
绛旓細lim(x鈫0)1-cos2x/鈭(0鈫sinx)ln(1+at)dt =lim(x鈫0)2x²/鈭(0鈫抯inx)ln(1+at)dt =lim(x鈫0)4x/ln(1+asinx)*cosx =lim(x鈫0)4x/asinx =4/a =1 鎵浠 a=4
绛旓細鍏蜂綋鍥炵瓟濡傚浘锛氳繛缁嚱鏁帮紝涓瀹氬瓨鍦ㄥ畾绉垎鍜屼笉瀹氱Н鍒嗭紱鑻ュ湪鏈夐檺鍖洪棿[a,b]涓婂彧鏈夋湁闄愪釜闂存柇鐐逛笖鍑芥暟鏈夌晫锛屽垯瀹氱Н鍒嗗瓨鍦紱鑻ユ湁璺宠穬銆佸彲鍘汇佹棤绌烽棿鏂偣锛屽垯鍘熷嚱鏁颁竴瀹氫笉瀛樺湪锛屽嵆涓嶅畾绉垎涓瀹氫笉瀛樺湪銆
绛旓細涓嶇煡閬撲綘鏄粈涔堟枃鍖栧眰娆 濡傛灉鏄ぇ瀛 浣犲氨鍙互鐪嬫噦 limx瓒嬩簬0,x²/(1-cos2x) 锛堝垎瀛愩佸垎姣嶅垎鍒眰瀵硷級=limx瓒嬩簬0,(2x)/(2sin2x) 锛堢户缁垎瀛愬垎姣嶆眰瀵硷級=limx瓒嬩簬0,2/(4cos2x) ( 鍥犱负x瓒嬩簬0鏃 cos2x=1)=2/4 =1/2
绛旓細褰搙瓒嬩簬0鐨鏃跺欙紝1-cos2x鏄瓑浠蜂簬0.5*(2x)^2鍗2x^2鐨勶紝鎵浠ヤ笌x^2鐩告瘮搴旇鏄悓闃堕潪绛変环鏃犵┓灏 閫夋嫨绛旀C
绛旓細=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) 2sinx/x =2 ...
绛旓細鎵浠1-cos2x=2sinx^2 褰搙瓒嬩簬0鏃讹紝sinx锝瀤 鎵浠瓒嬩簬0鏃讹紝sinx^2锝瀤^2 鎵浠1鈥攃os2x绛変环鏃犵┓灏忔槸2x^2 鏋侀檺 鏁板鍒嗘瀽鐨勫熀纭姒傚康銆傚畠鎸囩殑鏄彉閲忓湪涓瀹氱殑鍙樺寲杩囩▼涓紝浠庢荤殑鏉ヨ閫愭笎绋冲畾鐨勮繖鏍蜂竴绉嶅彉鍖栬秼鍔夸互鍙婃墍瓒嬪悜鐨勬暟鍊(鏋侀檺鍊)銆傛瀬闄愭柟娉曟槸鏁板鍒嗘瀽鐢ㄤ互鐮旂┒鍑芥暟鐨勫熀鏈柟娉曪紝鍒嗘瀽鐨勫悇绉嶅熀鏈...
