X趋于0时,求(tanx-sinx)/x^3的极限为什么不用tanx~x和sinx~x 求极限x趋向于0，limtanx-sinx/x^3 为什么这...
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我学的是同济的高数教材，上面有等价无穷小替换的规则，你可以自己看看，简化地说，就是加减法里面不能用，乘除法里面可以随便换。至于为什么，请看证明过程就行了。
绛旓細璁緔=[(1+tanx)/(1+sinx)]^(1/sinx)lny=(1/sinx)ln[(1+tanx)/(1+sinx)]lim[x鈫0] lny=lim[x鈫0] (1/sinx)ln[(1+tanx)/(1+sinx)]=lim[x鈫0] (1/sinx)ln[(1+sinx-sinx+tanx)/(1+sinx)]=lim[x鈫0] (1/sinx)ln[1+(tanx-si...
绛旓細lim(X瓒嬪悜浜0锛夛紙X骞虫柟/1-XtanX/1) =lim(X瓒嬪悜浜0锛(tanx-x)/(x^2tanx) 鐢辨棤绌峰皬閲忎唬鎹 =lim(X瓒嬪悜浜0锛夛紙tanx-x)/(x^3) 涓婁笅姹傚 =lim(X瓒嬪悜浜0锛夛紙1/(cosx^2)-1)/(3x^2) 鍐嶆眰瀵 =lim(X瓒嬪悜浜0锛塠-2/锛坈osx锛塣3]*si...
绛旓細x~tanx~sinx 鐨勮繖涓叧绯 3x~tan 3x 娉ㄦ剰鍒 tan 锛90掳- x锛= 1/tan x ,tan[3*(90掳- x)]=tan[ 270掳-3x]= 1/ tan 3x 鎵浠ワ紝浠 t =90-x 鍘熷紡 = lim 锛坱an 3t / tan t锛 锛宼 瓒嬭繎浜庨浂 锛鍙互鐢ㄧ瓑浠锋棤绌峰皬浜 = 3t / t = 3 ...
绛旓細鐢遍鎰忥紝f(tanx)锛.sinx0?10cosx0sinx01.=cosx+sinxcosx浠anx=-1鈭磝=k蟺-蟺4鎴杧=k蟺+3蟺4鈭碿osx+sinxcosx=2?12鎴?2+12鍗筹細f锛-1锛=2?12鎴?2+12鏁呯瓟妗堜负锛2?12鎴?<
绛旓細鐢═alyor灞曞紑涓涓嬪氨鍙互鐪嬪嚭锛岃櫧鐒堕兘鏄瓑浠锋棤绌峰皬锛屼絾绛変环鏃犵┓灏忕殑宸笉涓瀹氭槸绛変环鐨勶紝姣斿锛歴inx=x-x^3/3+x^5/5!-鈥︹︿簬鏄痩im锛坰inx-x锛/x^3鍙笉鏄0浜嗐
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绛旓細(x鈮1) 鏃犻檺鍏紡 sinx=x(1-x^2/蟺^2)(1-x^2/4蟺^2)(1-x^2/9蟺^2)鈥︹ cosx=(1-4x^2/蟺^2)(1-4x^2/9蟺^2)(1-4x^2/25蟺^2)鈥︹ tanx=8x[1/(蟺^2-4x^2)+1/(9蟺^2-4x^2)+1/(25蟺^2-4x^2)+鈥︹ secx=4蟺[1/(蟺^2-4x^2)-1/(9蟺^2-4x^2)+1/(25蟺^...
绛旓細c/b+b/c>=2(鍩烘湰涓嶇瓑寮忥細a+b>=2鈭(ab) 锛夊綋涓斾粎褰揵=c鏃剁瓑鍙锋垚绔嬶紝鍗砨=c鏃讹紝c/b+b/c鏈夋渶澶у2.姝ゆ椂涓夎褰BC涓虹瓑鑵颁笁瑙掑舰锛屾墍浠=c=(鈭3/3)*a 鎵浠ョ敱浣欏鸡瀹氱悊锛屽緱 cosA=(b^2+c^2-a^2)/(2bc)锛濓紞1/2 鎵浠锛120搴 鎵浠=(1/2)*bc*sinA=(鈭3/12)a^2 ...
绛旓細sinx=x(1-x^2/蟺^2)(1-x^2/4蟺^2)(1-x^2/9蟺^2)鈥︹ cosx=(1-4x^2/蟺^2)(1-4x^2/9蟺^2)(1-4x^2/25蟺^2)鈥︹ tanx=8x[1/(蟺^2-4x^2)+1/(9蟺^2-4x^2)+1/(25蟺^2-4x^2)+鈥︹ secx=4蟺[1/(蟺^2-4x^2)-1/(9蟺^2-4x^2)+1/(25蟺^2-4x^2)-+鈥︹...
绛旓細(sinx)^2+(cosx)^2=1.(1)sinx/cosx=tanx=t.(2)鐢变簬t澶т簬0鎵浠鏄涓鎴栫涓夎薄闄愯鎵浠inx鍜宑osx鐨勫煎悓鏃朵负姝ｆ垨鍚屾椂涓鸿礋.(3)鐢(1)(2)(3)鍙緱 cosx=鏍瑰彿t^2+1鍒嗕箣1,sinx=鏍瑰彿t^2+1鍒嗕箣t鎴 cosx=璐熺殑鏍瑰彿t^2+1鍒嗕箣1,si...
