当x趋近于0.tanx-sinx除以x(1-cos2x) 当X趋近于0时，lim（tanx-sinx）除以limx的三...
lim(x->0\uff09=\uff08tanx-sinx)/(1-cos2x)\u5229\u7528\u7b49\u4ef7\u65e0\u7a77\u5c0f\u4ee3\u6362\uff0ctanx\uff5ex\uff0c1-cosx\uff5ex²/2\uff0c1-cosx\uff5e2x²\u3002lim(x->0\uff09=\uff08tanx-sinx)/(1-cos2x)=lim(x->0\uff09\uff08tanx-sinx)/(1-cos2x)=lim\uff08x\u21920\uff09tanx\uff081-cosx\uff09/\uff081-cos2x\uff09=lim\uff08x\u21920\uff09x³/\uff082x²\uff09=0
\u5c31\u662f\u7528\u6700\u76f4\u63a5\u6700\u7b80\u5355\u7684\u65b9\u6cd5\u505a,\u4e0b\u9762\u63d0\u4f9b2\u79cd\u4e0d\u540c\u7684\u65b9\u6cd5:\u53c2\u8003\u5b9a\u7406lim[x\u21920] sinx/x=1 lim[x\u21920] (tanx-sinx)/x�0�6
=lim[x\u21920] (sinx/cosx-sinx)/x�0�6
=lim[x\u21920] (sinx-sinxcosx)/(x�0�6cosx)
=lim[x\u21920] sinx(1-cosx)/(x�0�6cosx)
=lim[x\u21920] sin�0�6x(1-cosx)/(x�0�6sin�0�5xcosx)
=lim[x\u21920] (sinx/x)�0�6\u00b7(1-cosx)/(sin�0�5xcosx)
=lim[x\u21920] (sinx/x)�0�6\u00b7(1-cosx)/[(1-cos�0�5x)cosx]
=lim[x\u21920] (sinx/x)�0�6\u00b7(1-cosx)/[(1+cosx)(1-cosx)cosx]
=lim[x\u21920] (sinx/x)�0�6\u00b71/[(1+cosx)cosx]
=1\u00b71/(1+1)
=1/2 lim[x\u21920] (tanx-sinx)/x�0�6
=lim[x\u21920] (sinx/cosx-sinx)/x�0�6
=lim[x\u21920] (sinx/x)\u00b7(1-cosx)/(x�0�5cosx)
=lim[x\u21920] (sinx/x)\u00b7[1-(1-2sin�0�5x/2)]/(x�0�5cosx)
=lim[x\u21920] (sinx/x)\u00b72sin�0�5(x/2)/(x�0�5cosx)
=lim[x\u21920] (sinx/x)\u00b72sin�0�5(x/2)/[4(x/2)�0�5cosx]
=lim[x\u21920] (1/2)\u00b7(sinx/x)\u00b7[sin(x/2)/(x/2)]�0�5\u00b7(1/cosx)=1/2\u00b71\u00b71�0�5\u00b71=1/2
1/4
x->0
tanx = x+(1/3)x^3+o(x^3)
sinx = x-(1/6)x^3+o(x^3)
tanx - sinx = (1/2)x^3 +o(x^3)
1- cos2x
= 1 -[ 1 -(1/2)(2x)^2 +o(x^2)]
= 2x^2 +o(x^2)
x(1- cos2x ) = 2x^3 +o(x^3)
lim(x->0) (tanx-sinx)/[x(1- cos2x )]
=lim(x->0) (1/2)x^3/ (2x^3)
=1/4
f=cosx/√(1-sin2x)+sinx/√(1-cos2x)+tanx/√tan2x
定义域tan2x>0 →kπ/2<x<kπ/2+π/4
f最小正周期2π
x∈(0,2π)一个周期内
f₁=cosx/(cosx-sinx)+sinx/√2·sinx+√(tanx-tan³x)/2 0<x<π/4
单调递增，f>1+√2/2
f₂=cosx/(sinx-cosx)+sinx/√2·sinx-√(tan³x-tanx)/2 π/2<x<3π/4
单调递增，f<√2/2-1/2
f₃=cosx/(sinx-cosx)-sinx/√2sinx+√(tanx-tan³x)/2 π<x<5π/4
存在极大值点，极大值<0
f₄=cosx/(cosx-sinx)-sinx/√2sinx-√(tan³x-tanx)/2 3π/2<x<7π/4
单调递增，f<1/2-√2/2<0
∴f∈(-∞,√2/2-1/2)∪(√2/2+1/2,+∞)
绛旓細=sinx(1/cosx-1)=sinx(1-cosx)/cosx =sinx(1-cosx) x--->0 =x*(-0.5x^2) x--->0 =-0.5x^3 tanx-sinx鏄瘮x^2楂橀樁鐨勬棤绌峰皬閲,鍗硉anx-sinx=o锛坸^2锛
绛旓細姹傛瀬闄愶紝鏍规嵁鏋侀檺鎯呭喌鏉ヨ鏄庨棶棰樸傚叿浣撹В绛斿鍥炬墍绀
绛旓細sinx=2sin(x/2)cos(x/2)/(sin^2(x/2)+cos^2(x/2)),鍒嗗瓙鍒嗘瘝鍚岄櫎浠os^2(x/2),寰楀埌sinx=2tan(x/2)/(1+tan^2(x/2))tanx-sinx=2tan(x/2)/(1+tan^2(x/2))-2tan(x/2)/(1+tan^2(x/2)),閫氬垎鍚庯紝寰楀嚭tanx-sinx=4tan^3(x/2)/(1-tan^4(x/2))褰搙瓒嬭繎浜0鏃...
绛旓細tanx - sinx 鈭 (x锛媥³/3) - (x - x³/6)锛漻³/2锛屾墍浠 a锛1/2锛宬锛3銆
绛旓細褰 x 瓒嬭繎浜 0 鏃讹紝鎴戜滑鍙互璁＄畻 tan(x) 鍜 sin(x) 鐨勮繎浼煎硷紝鐒跺悗璁＄畻瀹冧滑鐨勫樊鍊笺備娇鐢ㄦ嘲鍕掔骇鏁板睍寮锛屾垜浠彲浠ュ緱鍒帮細tan(x) 鈮 x + (1/3)x^3 + (2/15)x^5 + ...sin(x) 鈮 x - (1/6)x^3 + (1/120)x^5 - ...鍥犳锛宼an(x) - sin(x) 鍙互杩戜技琛ㄧず涓猴細tan(x) -...
绛旓細1/4
绛旓細tanx锛峴inx=tanx(1-cosx)=tanx*sin²x/2,tanx~x,sinx~x 鎵浠ユ槸1+2=3闃
绛旓細鍋氭瘮 鍐嶇敤娲涘繀杈炬硶鍒 绠楀嚭鏉ユ槸涓父鏁帮紝鎵浠ユ槸鍚岄樁鏃犵┓灏
绛旓細x鈫0鏃tanx-sinx=tanx(1-cosx)=2tanx[sin(x/2)]^2,涓2x*(x/2)^2=x^3/2鏄瓑浠锋棤绌峰皬閲,鈭磒=3.
绛旓細姹傚閮借兘姹傛鏉ワ紝锛1+x骞虫柟锛夊垎涔嬩竴鏄痑rctan x 鐨勫鏁 杩樻湁锛岀瓑浠锋棤绌峰皬鍦ㄥ姞鍑忕殑鏃跺欎笉瑕佷贡鐢紝浣犺繖閲岀敤绛変环鏃犵┓灏忚挋瀵逛簡锛屽叾浠栧湴鏂瑰彲涓嶄竴瀹氬锛屽洜涓虹宸х瓑浠锋棤绌峰皬涔嬪悗锛寈骞虫柟椤圭殑绯绘暟涓嶅悓锛屾墠寰楀埌姝ｇ‘鐨勭粨鏋溿傜被浼奸鐩殑娲涘繀杈惧拰娉板嫆瑙ｇ瓟 ...
