limx趋近于0tanx-sinx/x3用等价无穷小替换求极限 用等价无穷小解决极限问题 当x→0时 tanx-sinx除以...
lim(x\u21920)(tanx-sinx)/x³ \u5229\u7528\u7b49\u4ef7\u65e0\u7a77\u5c0f\u6c42\u6781\u9650\u56e0\u4e3a\u65e0\u7a77\u5c0ftanx-sinx=x^3/2\uff0c\u6781\u9650=0.5
\u539f\u5f0f=lim\uff08sinx/cosx-sinx)/sin³x
\u7ea6\u5206
=lim(1/cosx-1)/sin²x
=lim(1-cosx)/(sin²xcosx)
1-cosx~x²/2
sinx~x
\u6240\u4ee5\u539f\u5f0f=(x²/2)/(x²cosx)
=1/(2cos0)
=1/2
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lim锛坰inx-x锛/x^3鍙笉鏄0浜嗐
绛旓細鍗曠函鐨勭瓑浠锋棤绌峰皬渚嬪瓙鏄細x->0 limtanx/x lim(1-cosx)/x lim(arctanx/x)鍧囧彲
绛旓細濡傛灉鐩存帴浠も娍x鈫0锛屾槸涓嶈兘瀵煎嚭瀵煎嚱鏁扮殑锛屽繀椤昏涓涓緟鍔╃殑鍑芥暟尾锛漚^鈯縳-1閫氳繃鎹㈠厓杩涜璁＄畻銆傜敱璁剧殑杈呭姪鍑芥暟鍙互鐭ラ亾锛氣娍x=loga(1+尾)銆傛墍浠(a^鈯縳-1)/鈯縳锛澪/loga(1+尾)=1/loga(1+尾)^1/尾 鏄剧劧锛屽綋鈯縳鈫0鏃讹紝尾涔熸槸瓒嬪悜浜0鐨勩傝lim尾鈫0(1+尾)^1/尾=e,鎵浠im尾鈫01/...
绛旓細= lim sin6x/sin2x = lim 6cos6x/2cos2x = -6 / -2 = 3 2.鐢辨礇姣旇揪娉曞垯 lim lncotx/lnx =lim -x/(sin^2xcotx)鐢辩瓑浠锋棤绌峰皬x~sinx 涓婂紡=lim -tanx/sinx=lim -1/cosx=-1 鍙傝冭祫鏂欙細http://zhidao.baidu.com/question/47307878.html?si=1 ...
绛旓細鐢变簬褰搙鈫0鏃讹紝1-cosx锝12x2锛宻inx锝瀤锛屽洜姝わ細limx鈫0(1−cosx)[x−ln(1+tanx)]sin4x=limx鈫012x2[x−ln(1+tanx)]x4=limx鈫0x−ln(1+tanx)2x2=limx鈫01−sec2x1+tanx4x娲涘繀杈炬硶鍒欙級=limx鈫01+tanx−sec2x4x(1+tanx)=...
