求下列数列的极限 求下列数列的极限，高等数学

\u6c42\u4e0b\u5217\u6570\u5217\u7684\u6781\u9650 \u8981\u6709\u8fc7\u7a0b

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\u4e00\uff1a\u6709\u9898\u76ee\u77e5\u8be5\u5f0f\u5b50\u6ee1\u8db3\u4f7f\u7528\u8bfa\u5fc5\u8fbe\u6cd5\u5219\u7684\u6761\u4ef6\uff0c\u56e0\u6b64\uff0c\u5bf9\u51fd\u6570f\u5206\u5b50\u5206\u6bcd\u5206\u522b\u6c42\u4e00\u9636\u5012\u6570\u5f97\u52304X^3/(3X^2)=4X/3 \u7531\u4e8eX\u8d8b\u4e8e1\uff0c\u6545\u6781\u9650\u4e3a4/3

\u4e8c\uff1a\u5c06\u88ab\u79ef\u51fd\u6570\u5206\u5b50\u5206\u6bcd\u540c\u65f6\u4e58\u4ee5\u221a(x+\u0394x) + \u221a(x \u5316\u7b80\u4e4b\u540e\u4ee3\u5165\u0394x=0\u5f97\u5230\u6781\u9650\u4e3a1/(2\u221ax)

\u4e09\uff1a\u672c\u9898\u76ee\u4e0e\u7b2c\u4e8c\u9898\u4e00\u6837\uff0c\u5148\u5c06\u51fd\u6570\u5206\u5b50\u5206\u6bcd\u540c\u4e58\u4ee5\u221a(2x+1) + 3\u5316\u7b80\u4e4b\u540e\u4ee3\u5165x=4\u5f97\u5230\u6781\u9650\u4e3a\u65e0\u7a77\u5927

\u56db\uff1a\u672c\u9898\u76ee\u4e0e\u4e8c\u3001\u4e09\u89e3\u6cd5\u4e00\u6837\uff0c\u5c06\u51fd\u6570\u5206\u5b50\u5206\u6bcd\u540c\u4e58\u4ee51+\u221a(tanx+1\uff09\u5316\u7b80\u4e4b\u540e\u4ee3\u5165x=0\u5f97\u5230\u6781\u9650\u4e3a-2

\u516d\uff1a\u5206\u5b50\u5206\u6bcd\uff08\u8fd9\u91cc\u5c06\u5206\u6bcd\u770b\u505a1\uff09\u540c\u65f6\u4e58\u4ee5\u221a(x^2+x+1) + \u221a(x^2-x+1)\u5c06\u5f97\u5230\u7684\u7ed3\u679c\u5316\u7b80\uff0c\u5316\u7b80\u540e\u5f97\u52302x / ( \u221a(x^2+x+1)+\u221a(x^2-x+1) )\u5c06\u5f0f\u5b50\u5206\u5b50\u5206\u6bcd\u540c\u65f6\u9664\u4ee5x\uff0c\u5206\u5b50\u53d8\u4e3a2\uff0c\u5206\u6bcd\u53d8\u4e3a\u221a(1 + 1/x +1/x^2 ) + \u221a(1 - 1/x + 1/x^2) \u56e0\u4e3ax\u4e3a\u65e0\u7a77\u5927\uff0c\u6240\u4ee5\u5f0f\u5b50\u221a(1 + 1/x +1/x^2 ) + \u221a(1 - 1/x + 1/x^2)\u4e2d\u6240\u6709\u542b\u6709x\u7684\u9879\u5747\u8d8b\u4e8e0\uff0c\u6b64\u65f6\u6211\u4eec\u53ef\u4ee5\u76f4\u63a5\u5c06\u5176\u89c6\u4e3a0\uff0c\u5f97\u5230\u221a(1 + 1/x +1/x^2 ) + \u221a(1 - 1/x + 1/x^2)\u8d8b\u4e8e2\uff0c\u6240\u4ee5\u6574\u4e2a\u6781\u9650\u4e3a1/2

\u4e03\uff1a\u56e0\u4e3asinx\u51fd\u6570\u6709\u754c\uff0c\u5f53x\u8d8b\u4e8e0\u65f6\uff0c\u5c31\u53ef\u4ee5\u7528\u65e0\u7a77\u5c0f\u4e58\u4ee5\u4e00\u4e2a\u6709\u754c\u51fd\u6570\u7684\u503c\u4ecd\u4e3a\u65e0\u7a77\u5c0f\u8fd9\u4e00\u7ed3\u8bba\u6765\u505a\uff0c\u5373\u672c\u9898\u6781\u9650\u4e3a0

\u516b\uff1a\u8fd9\u79cd\u9898\u76ee\u5148\u5c06\u5206\u6bcd\u4e58\u4ee52\uff08\u518d\u5c06\u6574\u4e2a\u51fd\u6570\u4e58\u4ee50.5\u5c31\u53ef\u4ee5\u4fdd\u8bc1\u51fd\u6570\u503c\u548c\u539f\u6765\u4e00\u6837\uff09\uff0c\u5f97\u5230\u7684\u5f62\u5f0f\u6b63\u662f\u9ad8\u6570\u8bfe\u672c\u4e0a\u9762\u7684\u6807\u51c6\u5f62\u5f0f\uff1a\u5f53x\u8d8b\u4e8e0\u65f6\uff0csinx/x\u7684\u6781\u9650\u4e3a1\uff0c\u7c7b\u4f3c\uff0c\u6240\u4ee5\u6781\u9650\u4e3a0.5

\u4e5d\uff1a\u5f53x\u8d8b\u4e8e0\u65f6\uff0c1-cosx\u53ef\u4ee5\u89c6\u4e3ax^2/2\u7684\u9ad8\u9636\u65e0\u7a77\u5c0f\uff0ctanx\u4e3ax\u7684\u9ad8\u9636\u65e0\u7a77\u5c0f\uff0c\u5373\u5206\u5b50\u53d8\u4e3ax^2/2\uff0c\u5206\u6bcd\u53d8\u4e3ax^2\uff0c\u6240\u4ee5\u6781\u9650\u4e3a1/2

\u5341\uff1a\u56e0\u4e3ax\u8d8b\u4e8e0\u65f6\uff0c2x\u4e5f\u8d8b\u4e8e0\uff0c\u6545tan2x\u53ef\u4ee5\u770b\u505a2x\uff0c\u5219\u51fd\u6570\u7684\u5206\u6bcd\u5c31\u53ef\u4ee5\u5199\u62102x\uff0c\u518d\u5c06\u53d8\u6362\u540e\u7684\u51fd\u6570\u5206\u5b50\u5206\u6bcd\u540c\u65f6\u4e58\u4ee5\u221a(1+x) + 1\u5f97\u5230\u6781\u9650\u4e3a1/2

\u5341\u4e00\uff1a\uff082x+1\uff09/(2x-1)\u5316\u7b80\u4e3a1+ 1/(x-0.5)\uff0c\u4ee4t= x-0.5 (x\u8d8b\u4e8e\u65e0\u7a77\uff0c\u6545t\u4e5f\u8d8b\u4e8e\u65e0\u7a77)\uff0c\u5373\u6709x=t+0.5 \uff0c\u6240\u4ee5\u539f\u51fd\u6570\u53ef\u4ee5\u5199\u4e3a \uff081+ 1/t\uff09\u7684\uff08t+0.5\uff09\u6b21\u65b9\uff0c\u53ef\u4ee5\u5199\u4e3a\uff1a\uff081+ 1/t )^t \u4e58\u4ee5\uff081+ 1/t\uff09^0.5 \uff0c\u5f53t\u8d8b\u4e8e\u65e0\u7a77\u65f6\uff0c\uff081+ 1/t )^t \u7684\u6781\u9650\u4e3ae\uff0c\uff081+ 1/t\uff09^0.5 \u7684\u6781\u9650\u4e3a1\uff0c\u5c06\u4e24\u4e2a\u6781\u9650\u76f8\u4e58\uff0c\u5c31\u5f97\u5230\u8981\u6c42\u7684\u6781\u9650\uff0c\u7ed3\u679c\u4e3ae

\u5341\u4e8c\uff1a\u5df2\u77e5x\u8d8b\u4e8e\u65e0\u7a77\u65f6\uff0c\uff081+1/x\uff09\u7684x\u6b21\u65b9\u7684\u6781\u9650\u4e3ae\uff08\u8bfe\u672c\u4e0a\u6709\uff09\uff0c\u56fa\u6709\uff081-1/x\uff09\u7684x\u6b21\u65b9\u5f53x\u8d8b\u4e8e\u65e0\u7a77\u7684\u6781\u9650\u4e3a1/e\uff0c\uff08\u6240\u6709\u8fd9\u79cd\u7c7b\u578b\u7684\u9898\u76ee\u90fd\u7528\u8fd9\u79cd\u89e3\u6cd5\uff09\u3002\u800c\u4e14\u539f\u51fd\u6570\u7684\u6781\u9650\u53ef\u4ee5\u770b\u505a\u662f\uff081-1/x\uff09\u7684x\u6b21\u65b9\u7684\u6781\u9650\u7684k\u6b21\u65b9\uff08\u6781\u9650\u7684\u6027\u8d28\u4e4b\u4e00\uff09\uff0c\u6545\u6781\u9650\u4e3a1/e^k

\u6536

(1)
a1=1<3，假设对于k∈N*，1≤ak<3，则a(k+1)=3- 1/ak
1/3<1/ak≤1，2<3- 1/ak<8/3，a(k+1)>0
k为任意正整数，因此对于任意正整数n，an恒>0
令lim an=A，(其中，A>0)
n→∞
A=3- 1/A
A²-3A+1=0
A²-3A+ 9/4=5/4
(A- 3/2)²=5/4
A>0，A=(3+√5)/2
lim an=(3+√5)/2
n→∞
(2)
a1=2，0<2<3，假设对于k∈N*，0<ak<3，则a(k+1)=1/(3-ak)
0<3-ak<3，a(k+1)>0
k为任意正整数，因此对于任意正整数n，an恒>0
令lim an=A
n→∞
A=1/(3-A)
3A-A²=1
A²-3A+1=0
A>0，A=(3+√5)/2
lim an=(3+√5)/2
n→∞

姹備笅鍒楁暟鍒楃殑鏋侀檺
绛旓細a1=1<3锛屽亣璁惧浜巏鈭圢*锛1鈮k<3锛屽垯a(k+1)=3- 1/ak 1/3<1/ak鈮1锛2<3- 1/ak<8/3锛宎(k+1)>0 k涓轰换鎰忔鏁存暟锛屽洜姝ゅ浜庝换鎰忔鏁存暟n锛宎n鎭>0 浠im an=A锛(鍏朵腑锛孉>0)n鈫掆垶 A=3- 1/A A²-3A+1=0 A²-3A+ 9/4=5/4 (A- 3/2)²=5/4 A...

姹備笅鍒楁暟鍒楃殑鏋侀檺
绛旓細鍥炵瓟锛(1) lim(n-> 鈭)(1+2^n+3^n+4^n)^(1/n) let L = lim(x->鈭)(1+2^x+3^x+4^x)^(1/x) lnL = lim(x->鈭)ln(1+2^x+3^x+4^x) / x ( 鈭/鈭) = [(ln2).2^x+(ln3).3^x+(ln4).4^x]/(1+2^x+3^x+4^x) = [(ln2).(2/4)^x+(ln3...

姹備笅鍒楁暟鍒楃殑鏋侀檺:lim(n鈫掆垶)
绛旓細1 褰揳锛瀊鏃 (a^n-b^n)/(a^n+b^n)=[1-(b/a)^n]/[1+(b/a)^n]鎵浠im(n鈫掆垶) =1/1=1 2 褰揳=b鏃 (a^n-b^n)/(a^n+b^n)=0 鎵浠im(n鈫掆垶)=0 3 褰揳锛渂鏃 (a^n-b^n)/(a^n+b^n)=[(a/b)^n-1]/[(a/b)^n+1]鎵浠im(n鈫掆垶) =-1/1=-1 ...

姹備笅鍒楁暟鍒楃殑鏋侀檺
绛旓細1锛 Xn = 1/2^n 褰 n鈫掆垶 鏃讹紝 2^ n鈫掆垶 锛屾墍浠 1/2^ n鈫0銆傦紙灏辨槸璇达紝褰撲竴涓暟瓒嬩簬鏃犵┓澶ф椂锛屽畠鐨勫掓暟瓒嬩簬0锛2锛 Xn = 2+1/n^2 褰 n鈫掆垶 鏃讹紝 n^2 鈫掆垶 锛屾墍浠 1/n^2 鈫0.鎵浠 2 + 1/n^2 鈫 2 ...

姹備笅鍒楁暟鍒楃殑鏋侀檺
绛旓細瑙ｇ瓟濡備笅:

姹備笅鍒楁暟鍒楁瀬闄
绛旓細n瓒嬩簬鏃犵┓锛屾牴鍙烽噷闈㈣秼浜2n锛屽師寮忓寲涓簊in锛2n锛夐櫎浠1/N 锛屾棤绌峰皬姣旀棤绌峰皬锛屾礇蹇呰揪娉曞垯锛=cos锛2n锛夐櫎浠1/(N*N)=鏃犵┓澶 1 闄や互 鏃犵┓灏

姹備笅鍒楁暟鍒楁瀬闄
绛旓細濡傚浘

姹備笅鍒楁暟鍒楃殑鏋侀檺,瑕佽缁嗙殑瑙ｉ杩囩▼
绛旓細lim(n->inf)[3n^2+n]/[2n^2-1] = lim(n->inf)[3+1/n]/[2-1/n^2] = 3/2 銆愬綋鍒嗗瓙锛屽垎姣嶉兘鏄棤绌峰ぇ鏃躲傚垎瀛愶紝鍒嗘瘝鍚岄櫎浠ヤ竴涓棤绌峰ぇ鍥犲瓙銆備娇寰楀垎瀛愶紝鍒嗘瘝涓嚦灏戞湁1涓笉鍐嶆槸鏃犵┓澶с鏋侀檺灏卞嚭鏉ヤ簡銆傘憀im(n->inf)[(3n)^2+n]/[(2n)^2-1] = lim(n->inf)[9+1/n]/[...

姹備笅鍒楁暟鍒楃殑鏋侀檺銆(绗(8)棰) 杩囩▼鍐欒缁嗕竴鐐,璋㈣阿!
绛旓細1<=1+cos²n<=2,鎵浠 n鈭1<=n鈭1+cos²n<=n鈭2鐢卞す鎸ゅ畾鐞嗭紝鏋侀檺绛変簬1

瑙傚療涓嬪垪鏁板垪鐨鍙樺寲瓒嬪娍,鍐欏嚭鏋侀檺
绛旓細1銆佽繛缁垵绛夊嚱鏁帮紝鍦ㄥ畾涔夊煙鑼冨洿鍐姹傛瀬闄锛屽彲浠ュ皢璇ョ偣鐩存帴浠ｅ叆寰楁瀬闄愬硷紝鍥犱负杩炵画鍑芥暟鐨勬瀬闄鍊煎氨绛変簬鍦ㄨ鐐圭殑鍑芥暟鍊 2銆佸埄鐢ㄦ亽绛夊彉褰㈡秷鍘婚浂鍥犲瓙锛堥拡瀵逛簬0/0鍨嬶級3銆佸埄鐢ㄦ棤绌峰ぇ涓庢棤绌峰皬鐨勫叧绯绘眰鏋侀檺 4銆佸埄鐢ㄦ棤绌峰皬鐨勬ц川姹傛瀬闄 5銆佸埄鐢ㄧ瓑浠锋棤绌峰皬鏇挎崲姹傛瀬闄愶紝鍙互灏嗗師寮忓寲绠璁＄畻 6銆佸埄鐢ㄤ袱涓瀬闄愬瓨鍦...