高数求极限答案,四道,100分 高数求极限,如图四道题,希望有具体解答过程
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参考
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绛旓細1.鐩存帴甯﹀叆 2.鎶撳ぇ澶存濇兂锛鏋侀檺涓洪浂 3.閲嶈鏋侀檺锛屽嵆e^10 4.绛変环鏇挎崲锛屽垎瀛愮瓑浠蜂簬-2x^2锛屽垎姣嶇瓑浠蜂簬x^2锛屽嵆-2
绛旓細绗竴涓 1/4 绗簩涓 0 绗笁涓-1 绗洓涓0
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绛旓細2銆俵 i m [(tanx-sinx)/sin3x]= l i m [(sinx/cosx-sinx)/sin3x]= l i m [(1/cosx-1)/(sinx)^2]x鈫0 x鈫0 x鈫0 = l i m {(1/cosx-1)/[1-(cosx)^2]}= l i m {1/[cosx(cosx+1)]}= 1/2 x鈫0 ...
绛旓細1.杩閬撻珮鏁版瀬闄愭眰瑙闂锛屾眰瑙杩囩▼瑙佷笂鍥俱2.姹傝В杩欓亾楂樻暟鏋侀檺闂锛屾眰瑙g粨鏋滅瓑浜2銆3.杩欓亾楂樻暟鏋侀檺姹傝В闂锛岃В鐨勭涓姝:灏嗗垎姣嶅厛绛変环銆傚嵆鍥句腑绗竴琛屻4.杩欓亾楂樻暟鏋侀檺姹傝В闂锛岃В鐨勭浜屾:灏嗗垎瀛愭湁鐞嗗寲銆傜瓑浠峰強鏈夌悊鍖栧悗锛屽緱鍥句腑绗簩琛屻5.杩欓亾楂樻暟鏋侀檺姹傝В闂锛岃В鐨勭涓夋:0/0鍨嬫瀬闄愰棶棰...
绛旓細鍥炵瓟锛氳繖鍑犻亾棰橀兘绗﹀悎1鐨勬棤绌峰ぇ娆℃柟杩欎竴鎯呭舰,鍥犳鍙互鐢ㄦ礇蹇呰揪娉曞垯鏉ユ眰銆
绛旓細lim(x鈥>1)x/(1-x)锛濃垶锛(2) lim(x鈥>1)(x²-1)/(x-1)²锛漧im(x鈥>1)(x+1)(x-1)/(x-1)²锛漧im(x鈥>1)(x+1)/(x-1)锛濃垶锛(3) lim(x鈥>1)[1/(1-x)-3/(1-x³)]锛漧im(x鈥>1)[(1+x+x²)/(1-x³)-3/(1-x³...
绛旓細鍒嗕韩涓绉嶈В娉曪紝搴旂敤骞夸箟浜岄」灞曞紑寮忓拰绛変环鏃犵┓灏忛噺鏇挎崲姹傝В銆傗埖(1+x)^伪=1+伪x+[伪(伪-1)/(2!)]x²+鈥︹︼紝鈭磝鈫0鏃讹紝(1+x)^伪=1+伪x+O(x²)~1+伪x銆傗埓鍘熷紡=lim(x鈫0)[(1+ax)-(1+bx)]/x=a-b銆備緵鍙傝冦