数列极限。第二题
\u4e24\u9053\u6570\u5217\u6781\u9650\u9898\u76ee2.\u9996\u5148a=0,\u5426\u5219\u6781\u9650\u4e3a\u65e0\u7a77\u5927\uff0c\u5c06\u5b83\u4ee3\u5165\u539f\u5f0f\uff0c\u518d\u5f97b=6.
9.\u5148\u4e58\u4ee51 - 1/2\uff0c\u8fde\u7eed\u5229\u7528\u5e73\u65b9\u5dee\u516c\u5f0f\uff0c\u518d\u9664\u4ee5 1 - 1/2\u3002\u7528a\u8868\u793a\u9898\u76ee\u6700\u540e\u90a3\u4e2a\u62ec\u53f7\u540e\u9762\u90a3\u4e2a\u6570\u3002
\u539f\u5f0f=2 lim (1- a²\uff09=2
D
\u7b49\u6bd4\u6570\u5217\uff5ban\uff5d\u4e2d\uff0c\u524dn\u9879\u548cSn\u6ee1\u8db3limSn=1/a1\u6709\u6781\u9650\u5b58\u5728 \u6240\u4ee5\u516c\u6bd4|q|<1
\u53c8\u56e0\u4e3a\u5728\u516c\u6bd4\u5c0f\u4e8e1\u7684\u7b49\u6bd4\u6570\u5217\u4e2d \u524dn\u9879\u548climSn=a1/(1-q)
\u6240\u4ee51/a1=a1/(1-q)
(a1)^2=1-q
\u56e0\u4e3aa1>1 |q|<1
\u6240\u4ee5a1\u7684\u53d6\u503c\u8303\u56f4\u662fD(1,\u6839\u53f72)
lim【n→∞】[2n+(an²-2n+1)/(bn+2)]
=lim【n→∞】{[2n(bn+2)+an²-2n+1]/(bn+2)}
=lim【n→∞】[(2bn²+4n+an²-2n+1)/(bn+2)]
=lim【n→∞】{[(a+2b)n²+2n+1]/(bn+2)}
这是一个∞/∞型的极限，应用洛比达法则，有：
lim【n→∞】{[(a+2b)n²+2n+1]/(bn+2)}
=lim【n→∞】{[2(a+2b)n+2]/b}
=2lim【n→∞】{[(a+2b)n+1]/b}
已知：lim【n→∞】[2n+(an²-2n+1)/(bn+2)]=1
即：2lim【n→∞】{[(a+2b)n+1]/b}=1
lim【n→∞】{[(a+2b)n+1]/b}=1/2
显然有：a+2b=0……………………(1)
代入极限，有：
lim【n→∞】(1/b)=1/2
1/b=1/2，解得：b=2
代入(1)，有：a+2×2=0，解得：a=-4
绛旓細=lim銆恘鈫掆垶銆憑[2n(bn+2)+an²-2n+1]/(bn+2)} =lim銆恘鈫掆垶銆慬(2bn²+4n+an²-2n+1)/(bn+2)]=lim銆恘鈫掆垶銆憑[(a+2b)n²+2n+1]/(bn+2)} 杩欐槸涓涓垶/鈭炲瀷鐨鏋侀檺锛屽簲鐢ㄦ礇姣旇揪娉曞垯锛屾湁锛歭im銆恘鈫掆垶銆憑[(a+2b)n²+2n+1]/(bn+2)} =lim...
绛旓細n(ln(n-1)-ln(n))=-nln(1+1/(n-1))=-ln(1+1/(n-1))^(n-1)-ln(1+1/(n-1))鈭磍im<n->锕⑩垶>n(ln(n-1)-ln(n))=-lne=-1
绛旓細绗竴棰楥鏋侀檺涓嶅瓨鍦紝濂囨暟涓1锛屽伓鏁颁负-1锛岃烦璺冿紝甯告暟鍒楄甯告暟灏辨槸鍏舵瀬闄愶紝0.3寰幆锛屾瀬闄愭槸1/3锛0.3*(1+1/10+鈥)=0.3*(1-(1/10)^n)/(1-1/10)=0.3/0.9
绛旓細=lim锛3+1/n锛/锛2+1/n锛=3/2锛宯瓒嬩簬鏃犵┓澶э紝鍏跺掓暟1/n瓒嬩簬0銆傛垨鎸夊畾涔墊锛3n+1锛/锛2n+1锛-3/2|=1/锛2n+1锛2<蔚锛屾绘湁N>锛1/蔚-2锛/4
绛旓細濡傚浘鎵绀
绛旓細=n娆℃牴鍙蜂笅(n)*A,鏋侀檺涓篈鐒跺悗灏嗚寮忕缉灏,a1,a2,...,am涓偗瀹氭湁涓涓拰A鐩哥瓑鐨,鎶婅繖涓椤圭暀涓,鍏朵綑椤瑰垹闄,杩欐牱灏辩缉灏忎簡,缁撴灉涓猴細n娆℃牴鍙蜂笅(A^n)=A鏀惧ぇ涓庣缉灏忓悗鐨勬瀬闄愰兘鏄疉,杩欐牱鐢卞す閫煎噯鍒,鏈寰楄瘉銆绗簩棰,棣栧厛瑕佽瘉鏄庢瀬闄愬瓨鍦,璇鏁板垪鍗曞鏄瘮杈冩樉鐒剁殑,涓嬮潰璇佹槑鏈夌晫,鏁板褰掔撼娉,x1銆
绛旓細鍥犱负xn鏈夌晫锛屾墍浠 瀛樺湪M>0锛屼娇|xn|鈮 鍙 limyn=0 鍗充换鎰徫>0,瀛樺湪N>0锛屽綋n>N鏃讹紝|yn|<蔚/M 浠庤 瀵逛簬鍒氭墠鐨凬,蔚,鏈墊xnyn|<M路蔚/M=蔚 鎵浠 鐢卞畾涔夛紝寰 limxnyn=0
绛旓細lim(x->0) [(1+x)^(1/2) +(1-x)^(1/2) -2]/x^2 =lim(x->0) -(1/4)x^2/x^2 =-1/4 (2)a1>1 a(n+1) = 2- 1/an {an} 鏄 閫掑噺鏁板垪 a(n+1) = 2- 1/an n>1 2- 1/an < an (an)^2 -2an + 1 >0 an< 1+2鈭2 c = max { a1, 1+2鈭2 ...
绛旓細2.(1)[(-1)鈭*1/(n+1)]=[1/(1+n)]锛溛碉紝鍙朜=[1/蔚-1],褰搉锛濶鏃讹紝瀵逛换鎰徫碉紴0锛屾湁 [(-1)鈭*1/(n+1)]锛溛垫垚绔嬨傛湁鐤戦棶璇疯拷闂紝婊℃剰璇烽噰绾硚\(鈮р柦鈮)/~
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