数列的极限不会求，希望能有大佬帮忙最好有计算过程 高数数列极限问题，求大佬帮忙🙏
\u6c42\u95ee\u8fd9\u9053\u6570\u5217\u6c42\u6781\u9650\u7684\u9898\u76ee\u600e\u4e48\u505a\uff1f\u5229\u7528: 2siny.cosy = sin2y
2sin(x/2^n).cos(x/2^n)
=sin[x/2^(n-1)]
2sin[x/2^(n-1)].cos[x/2^(n-1)]
=sin[x/2^(n-2)]
...
...
2sin(x/2).cos(x/2)
=sinx
(2) 原式=n→∞lim(2n^2-2+2)/(n^2-1)=n→∞lim[2(n-1)(n+1)+2]/[(n+1)(n-1)]
=n→∞lim2+1/(n+1)(n-1)=2
(3) 原式=n→∞lim(2n+2-1)/(n^2-1)=n→∞lim[2(n+1)-1]/(n+1)(n-1))=n→∞lim2/(n-1)=0
(4) 原式=n→∞lim(2n^3-2n+3n-3+3)/(n^2-1)=n→∞lim[2n(n+1)(n-1)+3(n-1)+2]/(n+1)(n-1)
=n→∞lim2n+3/(n+1)=n→∞lim2n+0=∞
n->∞时1/n和1/2^n的极限都为0，所以答案为1
原式 = 2(n^2 - 1)/(n^2 - 1) + 2/(n^2 - 1) = 2 + 2/(n^2 - 1),n->∞时后面一部分极限为0，所以答案为2.
原式 = 2(n + 1)/((n + 1)(n-1)) - 1/(n^2 - 1) = 2/(n - 1) - 1/(n^2 - 1),n->∞时，两部分都为0，所以答案为0.
题目太过模糊，看不清楚
绛旓細=n鈫掆垶lim2n+3/(n+1)=n鈫掆垶lim2n+0=鈭
绛旓細n!/n^n =(1/n)(2/n)...(n/n)鈮 [1/n +2/n+...+n/n]/n } ^n ={ [n(n+1)/2 ]/n^2 } ^n =[ (1/2)(1 + 1/n) ]^n lim(n->鈭) [ (1/2)(1 + 1/n) ]^n =0 0<n!/n^n <[ (1/2)(1 + 1/n) ]^n =>lim(n->鈭) n!/n^n =0 ...
绛旓細绗竴棰橈紝鍒嗗瓙鏄垎姣嶇殑楂橀樁鏃犵┓灏忥紝鎵浠ュ垎瀛愯偗瀹氫负0锛岀畻寰桝=2锛屽洖浠=2锛岀敤娲涘繀杈剧畻寰桞=1/2 绗簩棰橈紝Xn=1+X(n-1)/(1+X(n-1))锛岀敤棰橀噷缁欑殑鍑忎竴涓嬭繖涓紡瀛愬彂鐜版槸澶т簬0鐨勶紝鎵浠ヨ繖涓鏁板垪鍗曡皟閫掑銆備笖Xn蹇呯劧灏忎簬2锛屾墍浠ユ槸鏈夎В鐨勶紝閭ｄ箞褰搉瓒嬪悜浜庘垶鏃讹紝Xn+1=Xn锛岃В涓涓柟绋嬪氨琛屼簡銆倄...
绛旓細a1=1, a2=2 3a(n+2)-4a(n+1)+an =0 3a(n+2) - a(n+1) = 3a(n+1) -an => { 3a(n+1) -an } 鏄瓑宸鏁板垪, d=0 3a(n+1) -an = 3a2 -a1 = 6-1 =5 3a(n+1) -an = 5 a(n+1) = (1/3)an +5/3 a(n+1) -5/2 = (1/3)( an - 5/2)=...
绛旓細1銆佷护e(n)=1+1+1/2!+1/3!+...+1/(n+1)!锛屽垯e(n)鈥>e锛2銆佷护s(n)=1/2!+2/3!+...+n/(n+1)!锛屽垯s(n)+e(n)=1+1+2/2!+3/3!+...+(n+1)/(n+1)!=1+e(n-1)锛3銆佷护涓よ竟鐨刵瓒嬩簬鏃犵┓澶э紝鍒欏彲浠ュ緱鍒皊(n)鈥>1銆
绛旓細浠ヤ笂
绛旓細3)x^n-1=(x-1)(1+x+x^2+鈥︹+x^n-1) 锛堝鏋滀笉鐭ラ亾鍙互浠庣瓑姣鏁板垪1+x+x^2+鈥︹+x^n-1姹傚墠n椤瑰拰寰楀嚭锛夋墍浠ュ師鏋侀檺=lim(1+x+x^2+鈥︹+x^n-1)=n 4)lim(鈭(1+x)-鈭歺)=lim 1/(鈭(1+x)+鈭歺)=0 5)鍘熷紡=lim (1+x-3)/(1-x²)=lim (x-2)/(1-x)...
绛旓細2銆佹眰涓嬪垪鏁板垪鐨勬瀬闄 1锛夈乴im(2-1/n) = 2-0 =2 (n-->鈭)锛2锛夈佸師寮= 3-0+0 = 3 (n-->鈭)锛3锛夈佸師寮 =lim[(3-2/n+1/n^2)/(8/n^2-1)]=3/(-1)=-3 锛坣-->鈭烇紝鍒嗗瓙鍒嗘瘝鍚屾椂闄や互n^2锛4锛夈佸師寮=lim[(2-1/n+1/n^2)/(3+1/n^2)=2/3 锛坣-->鈭...
绛旓細姝ら姝ｇ‘锛屾瘮濡傛尟骞呬笉鏂噺灏忕殑娉㈠姩鍑芥暟锛屾瀬闄瀛樺湪锛屼絾鏄苟涓嶅崟璋冿紝鑰屾槸姝ｈ礋娉㈠姩鐨勩傛瀬闄愬瓨鍦紝鏍规嵁鏋侀檺瀹氫箟鍙煡锛屽繀鐒舵湁鐣屻傛棤鐣鏁板垪鎬绘槸瀛樺湪澶т簬缁欏畾鍊肩殑椤
绛旓細n/(2n^2+1)+ n/(2n^2+2)+...+n/(2n^2+n)n^2/(2n^2+n) 鈮/(2n^2)+1+ n/(2n^2+2)+...+n/(2n^2+n)鈮 n^2/(2n^2+1)lim(n->鈭) n^2/(2n^2+n) =lim(n->鈭) n^2/(2n^2+1) =1/2 => lim(n->鈭) [n/(2n^2+1)+ n/(2n^2+2)+...+n/...
