高数三角函数极限问题
\u9ad8\u6570\u4e09\u89d2\u51fd\u6570\u6781\u9650\u95ee\u9898(cosn)^2=(1+cos2n)/2,\u5f53n\u8d8b\u8fd1\u4e0e\u65e0\u7a77\uff0ccos2n\u7684\u503c\u4e0d\u786e\u5b9a\uff0c\u6781\u9650\u4e0d\u5b58\u5728
lim\uff08x\u8d8b\u5411\u516d\u5206\u4e4bpai\uff09\uff08sinx + 1\uff09/(sinx-1)
=(1/2+1)/(1/2-1)
=3
sin2x~2x，sin5x~5x（x→0）极限为2/5 cosx=sin（π/2-x）sin（π/2-x）~π/2-x（x→π/2）极限为-1
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绛旓細sin(nPi)=sin(0)锛屽緱鍑鏋侀檺涓0 sin(xpi)鏄病鏈夋瀬闄愮殑锛屽湪-1鍒1涔嬮棿鎽嗗姩
绛旓細lim(sinax/sinbx)=a/b (杩欏簲璇ヤ笉闅捐瘉鏄)1. =lim(x/sin9x)*lim(sin2x/sin9x)=1/9*2/9=2/812. =1/銆恖im(5x/sin3x)+lim(sinx/(cosx*sin3x))銆=1/锛5/3+1/3锛=1/2
绛旓細瑙ｏ細鍘熷紡 = [ x^2 *cos(x^2) - (sinx)^2 ]/ [ (x^2 * (sinx)^2 ] * [ 1/ cos (x^2 ) ]= { [ (x^2 - (1/2!) * x^4 +o( x^4) ] } - [ x^2 - (2/3!) * x^4 + o(x^4 ) ]/ [ (x^4 + o(x^4)] * [ 1/ cos (x^2 ) ]...
绛旓細鏋侀檺棣栧厛搴旇鑰冭檻鐨勬槸鑷彉閲忕殑鍙樺寲杩囩▼锛岀浜岋紝瑕佺悊瑙ｆ瀬闄愭椂涓涓‘瀹氱殑甯告暟锛屾槸涓涓暟銆傜劧鍚庤冭檻浣犺鐨涓夎鍑芥暟锛屽厛鐪媠in(x) 鍜宑os(x)锛屽綋鑷彉閲弜瓒嬩簬鏃犵┓澶ф椂锛屾瀬闄愪笉瀛樺湪銆俿in(x)褰撹嚜鍙橀噺x瓒嬩簬0鏃讹紝鏋侀檺涓0锛沜os(x)褰撹嚜鍙橀噺x瓒嬩簬0鏃讹紝鏋侀檺涓1銆倀an(x)褰撹嚜鍙橀噺x瓒嬩簬0鏃讹紝鏋侀檺涓0锛泃an(x)褰...
绛旓細鎵浠im x瓒嬪悜浜0锛(sin 1/x锛/(1/x)=0锛涘姹傝В渚嬪lim x瓒嬪悜浜庘垶鎯呭喌涓(sin 1/x锛/(1/x)锛1/x瓒嬪悜浜0锛宻in 1/x瓒嬪悜浜0锛屾鏃剁敱浜巐im x瓒嬪悜浜0锛(sin x锛/x=1锛屾墍浠im x瓒嬪悜浜庘垶锛(sin 1/x锛/(1/x)=1锛涘浜0/0锛屸垶/鈭烇紝鈭灺扁垶绛夌瓑鎯呭喌寰寰闇瑕佸彲浠ヤ娇鐢ㄥす閫煎畾鐞嗐佺綏蹇...
绛旓細arctgx路sin1/x x-->0鐨勬椂鍊檃rctgx-->0锛岃宻in1/x鏄竴涓湁鐣岀殑閲忥紝[-1,1]鎵浠ヤ袱鑰呬箣绉繕鏄秼杩戜簬0銆傚畠瀵逛簬2鏉ヨ鏄棤绌峰皬锛岀暐鍘 鎵浠ユ渶鍚庣瓟妗堝氨鏄 ln2
绛旓細鏍规嵁閲嶈鏋侀檺锛歭im(x->0)sin/x=1鏄撳緱锛歭im(x->0)x/sinx=lim(x->0)1/(sin/x)=1,鑰宭im(x->0)锛坈osx锛塣2=1^2=1,鎵浠ワ細lim(x->0)(cosx)^2/[1+(cosx)^2]=1/(1+1)=1/2,鑰宭im(x->0)2/(1+x^2)=2/[1+lim(x->0)x^2]=2,鎵浠ュ師鏋侀檺=1*(1/2)*2=1 鏋侀檺...
绛旓細鍚庨潰鏄痵inx-xcosx 姹傚锛岋紝鍓嶉潰姹傚涔嬪悗绛変簬 2cosx-xsinx = 2锛坸->0)鍚庨潰姹傚涔嬪悗绛変簬锛寈sinx 鍒嗘瘝鍒嗘垚涓ら儴鍒嗭紝x ,, x sin^2x 鐢变簬 鍓嶉潰閭ｉ儴鍒嗗凡缁忛浂姣旈浂褰㈠紡 锛屾墍浠ユ渶鍚庡甫鍏0 绛変簬2 锛2cos0 = 2*1 = 2)鍚庨潰閭ｉ儴鍒 xsinx/3x^2 (xsin^2x姹傚鐨勬潵鐨)鍥犱负x瓒嬩簬闆 鎵浠...
