(1-cosX)的X次方求极限用洛必达
lim[x-->0](1-cosx)^x先求其对数的极限
lim[x-->0]ln(1-cosx)^x
=lim[x-->0]xln(1-cosx)
=lim[x-->0]ln(1-cosx)/(1/x) ---“ ∞/∞“型，用洛必达法则
=lim[x-->0][-x^2*sinx]/[1-cosx]-----1-cosx~1/2x^2
=lim[x-->0][-x^2*sinx]/[1/2x^2]
=0
所以，lim[x-->0](1-cosx)^x=e^0=1
绛旓細鍏堟眰鍏跺鏁扮殑鏋侀檺 lim[x-->0]ln(1-cosx)^x =lim[x-->0]xln(1-cosx)=lim[x-->0]ln(1-cosx)/(1/x) ---鈥 鈭/鈭炩滃瀷锛岀敤娲涘繀杈炬硶鍒 =lim[x-->0][-x^2*sinx]/[1-cosx]---1-cosx~1/2x^2 =lim[x-->0][-x^2*sinx]/[1/2x^2]=0 鎵浠ワ紝lim[x-->0](1-...
绛旓細1-锛坈osx锛²绛変环浜巗in²x銆傜瓑浠锋棤绌峰皬鏄棤绌峰皬鐨涓绉嶃傜瓑浠锋棤绌峰皬鏇挎崲鏄绠楁湭瀹氬瀷鏋侀檺鐨勫父鐢ㄦ柟娉曪紝瀹冨彲浠ヤ娇姹傛瀬闄闂鍖栫箒涓虹畝锛屽寲闅句负鏄撱傛眰鏋侀檺鏃讹紝浣跨敤绛変环鏃犵┓灏忕殑鏉′欢锛氳浠ｆ崲鐨勯噺锛屽湪鍙栨瀬闄愮殑鏃跺欐瀬闄愬间负0锛涜浠ｆ崲鐨勯噺锛屼綔涓鸿涔樻垨鑰呰闄ょ殑鍏冪礌鏃跺彲浠ョ敤绛変环鏃犵┓灏忎唬鎹紝浣嗘槸浣滀负...
绛旓細杩欓亾棰樺叾瀹炴尯绠鍗曠殑锛屽氨鏄洿鎺ユ妸x=0浠ｈ繘鍘诲嵆鍙紝鍘熷紡=(1+cos0)鈭0=2鈭0=1 灏辫繖涔堢畝鍗曘傞噰绾充笌鍚︼紝璇峰悰鑷究
绛旓細x鐨1-cosx鏂圭殑鏋侀檺鈭1-cosx=1-{1-2sin?(x/2)}=2sin?(x/2)鍙堚埖sin(x/2)涓(x/2)鏄瓑浠锋棤绌峰皬鈭2sin?(x/2)涓2*(x/2)?鍗(x?)/2鏄瓑浠锋棤绌峰皬鈭1-cosx鐨勬瀬闄绛変簬(x?)/2鐨勬瀬闄愩傚崟璋冩湁鐣屽噯鍒欙細鍗曡皟澧炲姞锛堝噺灏戯級鏈変笂锛堜笅)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...
绛旓細鐢辨鍙锛宯鈮0鏃讹紝灏5鐨勶紙n+1锛夋鏂鍙樹负5鐨刵娆℃柟闇闄や互涓涓5锛屾墍浠ュ彲瀹氫箟5鐨0娆℃柟涓猴細5 梅 5 = 1 1-鈭cosx鐨绛変环鏃犵┓灏忥細x^2/4銆傚垎鏋愯繃绋嬪涓嬶細鍒╃敤cosx=1锛峹^2/2+o(x^2)=1锛(1+cosx锛1)^鎭掔瓑鍙樺舰=1锛(1+(cosx锛1)/2)+o(cosx锛1)锛漻^2/4+o(x^2)銆姹傛瀬闄鏃讹紝浣跨敤...
绛旓細lim(1-cosx)/x^2(x瓒嬩簬0锛=1/2銆傝В绛旇繃绋嬪涓嬶細鈥鏋侀檺鈥濇槸鏁板涓殑鍒嗘敮鈥斺斿井绉垎鐨勫熀纭姒傚康锛屽箍涔夌殑鈥滄瀬闄愨濇槸鎸団滄棤闄愰潬杩戣屾案杩滀笉鑳藉埌杈锯濈殑鎰忔濄傛暟瀛︿腑鐨勨滄瀬闄愨濇寚锛氭煇涓涓嚱鏁颁腑鐨勬煇涓涓彉閲忥紝姝ゅ彉閲忓湪鍙樺ぇ锛堟垨鑰呭彉灏忥級鐨勬案杩滃彉鍖栫殑杩囩▼涓傞愭笎鍚戞煇涓涓‘瀹氱殑鏁板糀涓嶆柇鍦伴艰繎鑰屸...
绛旓細浣犺繖鏍峰仛鏄笉琛岀殑銆
绛旓細鍏蜂綋鍥炵瓟濡傚浘锛氬厛瑕佺敤鍗曡皟鏈夌晫瀹氱悊璇佹槑鏀舵暃锛岀劧鍚庡啀姹傛瀬闄鍊笺傚簲鐢ㄥす鎸ゅ畾鐞嗙殑鍏抽敭鏄壘鍒版瀬闄愬肩浉鍚岀殑鍑芥暟锛屽苟涓旇婊¤冻鏋侀檺鏄秼浜庡悓涓鏂瑰悜 锛屼粠鑰岃瘉鏄庢垨姹傚緱鍑芥暟鐨勬瀬闄愬笺
绛旓細lim(x->0+) (1-cosx)*lnx =lim(x->0+) (x^2/2)*lnx =(1/2)*lim(x->0+) lnx/(1/x^2)=(1/2)*lim(x->0+) (1/x)/(-2/x^3)=(-1/4)*lim(x->0+) x^2 =0
绛旓細杩欎釜棰樼洰寰堟槑鏄句簡锛1-(cosx)^2=(sinx)^2 x瓒嬩簬0 鎵浠ワ紙sinx锛塣2鍙互鏃犵┓灏忔浛鎹㈡垚x^2 缁撴灉鏄1 涓夎鍑芥暟棰樹綘瑕侀氳繃閭ｄ簺閲嶈鐨勮瀵煎叕寮忓彉褰㈡垚鍙互鏃犵┓灏忔浛鎹㈢殑褰㈠紡鐒跺悗璁＄畻 鑰屾湁鐨勭畝鍗曠殑灏辨槸閲嶈鐨勬棤绌峰皬鏇挎崲鐨勫舰寮 鎵浠ョ粓褰掓潵璇寸啛璁伴噸瑕佹棤绌峰皬绛変环鍜岃瀵煎叕寮 ps锛氫笅闈㈤偅浣嶅ぇ鍝ヤ綘娲涘繀杈炬眰瀵奸兘姹傞敊浜...
