一道极限的题，谢谢大家
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\u6240\u4ee5\u8981\u51d1\u6210lim\uff08x\u2192\u221e\uff09\uff08\uff081+1/x\uff09^x\u8fd9\u79cd\u5f62\u5f0f\uff0c
\u4e0b\u9762\u7684x\u7edf\u4e00\u6362\u6210n\u5c31\u53ef\u4ee5\u4e86\uff0c\u8fd8\u6709\u54ea\u4e00\u6b65\u4e0d\u660e\u767d\uff0c
\u53ef\u4ee5\u7ed9\u6211\u7559\u8a00\uff0c\u6211\u4e0a\u7ebf\u540e\u5e2e\u4f60\u89e3\u7b54\uff01
lim\uff08x\u2192\u221e\uff09[\uff082n+3\uff09/(2n+1)]^(n+1)
=lim\uff08x\u2192\u221e\uff09[\uff082n+1+2\uff09/(2n+1)]^(n+1)
=lim\uff08x\u2192\u221e\uff09(1+2/(2n+1))^(n+1)
=lim\uff08x\u2192\u221e\uff09[\uff081+2/(2n+1))^(2n+1)/2]^[2(n+1)/(2n+1)]
=lim\uff08x\u2192\u221e\uff09[\uff081+2/(2n+1))^(2n+1)/2]^lim\uff08x\u2192\u221e\uff09[2(n+1)/(2n+1)](\u6781\u9650\u8fd0\u7b97\u6cd5\u5219\uff1a\u5e42\u7684\u6781\u9650\u7b49\u4e8e\u6781\u9650\u7684\u5e42)
=e^lim\uff08x\u2192\u221e\uff09[1+1/(2n+1)]
=e^[lim\uff08x\u2192\u221e\uff091+lim\uff08x\u2192\u221e\uff091/(2n+1)]
=e^(1+0)
=e
\u6709\u7591\u95ee\uff0c\u53ef\u4ee5hi\u6211\u3002
\u70b9\u51fb\u770b\u5927\u56fe
数列首项为1 公比为1/2
根据等比数列的求和公式 1+1/2+......+1/(2^n)=1*[1-(1/2)^n]/(1-1/2)=2-2*(1/2)^n
当n趋近于无穷大时，(1/2)^n就趋近于0 所以 数列的和就趋近于2
即数列的极限为2
解答完毕 希望对楼主有帮助
根据等比数列求和,1+1/2+1/4+...+(1/2)^n=2-2(1/2)^(n+1),故当n趋向无穷大时,1+1/2+1/4+...+(1/2)^n=2.
2
先用求和公式求出表达式为2-（1/2）^n，当n趋于无穷时，极限为2
绛旓細鍥炵瓟锛歭im(x->鈭) (x^2-1)/(1-x)鏄垶/鈭炲瀷鏋侀檺,杩愮敤娲涘繀濉旀硶鍒欒绠: lim(x->鈭) (x^2-1)'/(1-x)'=lim(->鈭)2x/-1 =-鈭 鎴栬呭埄鐢ㄥ洜寮忓垎瑙(x^2-1)/(1-x) =(x+1)(x-1)/(1-x)=-(x+1) 鍙互鍚屾牱寰楀埌缁撹:鎵姹傛瀬闄愪笉瀛樺湪銆
绛旓細(1) lim(x鈫1)(x^2-2x+1)/(x^2-1)=lim(x鈫1)(x-1)^2/[(x-1)(x+1)]=lim(x鈫1)(x-1)/(x+1)=0 (2) lim(x鈫4)(x^2-6x+8)/(x^2-5x+4)=lim(x鈫4)(x-2)(x-4)/[(x-1)(x-4)]lim(x鈫4)(x-2)/(x-1)=2/3 (3) 鍘熷紡=lim(x鈫2)(x+2)...
绛旓細=lim(x->0+)(lntan2x)/(1/x)=lim(x->0+)(1/tan2x)*sec²2x*2/(-1/x²)=-2lim(x->0+)x²/tan2x =0 鎵浠 鍘熷紡=e^0=1
绛旓細姹傞珮鏁板ぇ绁炲府蹇欒В绛涓閬撴眰鏋侀檺鐨勯锛侊紒璋㈣阿锛侊紒鍏蜂綋瑙ｇ瓟濡傚浘鎵绀
绛旓細鍒: lny=x*ln(1+(x/a))(1/y)*y'=ln(1+(x/a))+x*(1/(1+(x/a)))*(1/a)=ln(1+(x/a))+(x/(x+a))y'=y*[ln(1+(x/a))+(x/(x+a))][(1+(x/a))^x]'=[(1+(x/a))^x]*[ln(1+(x/a))+(x/(x+a))]鍘熷紡 =lim[(1+(x/a))^x-1]/x^2 * ...
绛旓細杩欐槸涓涓 0/0 鍨嬬殑鏋侀檺锛鍙互浣跨敤缃楀繀濉旀硶鍒欙細=lim[½ * (-1)/鈭(3-x) - ½ * 1/鈭(1+x)]/(2x)=lim[½ * (-1)/鈭2 - ½ * 1/鈭2]/(2*1)=lim [(-1)/鈭2]/2 =-1/(2鈭2)
绛旓細鍒嗗瓙鏈夌悊鍖栵紝鍒嗗瓙鍒嗘瘝鍚屼箻浠((x^2+6x+5)^(1/2) + x锛夊師寮=lim((x^2+6x+5) - x^2)/((x^2+6x+5)^(1/2) + x锛=lim(6x+5)/((x^2+6x+5)^(1/2) + x锛 (x瓒嬭繎浜庢鏃犵┓)涓婂紡鐨勫垎瀛愬垎姣嶅悓闄や互 x =lim(6+ 5x)/((1+ 6/x + 5/x^2)^(1/2) + 1锛 (x瓒嬭繎...
绛旓細绗竴涓紝褰搙<1鏃讹紝(x^n)^2<x^n锛屽緱鍑烘瀬闄愪负x 绗簩涓紝鏉′欢閿欒锛屽簲璇ヤ负x>1锛寈^(2n)>>x^n锛屽緱鍑烘瀬闄愪负x^2 褰搙=1锛屾瀬闄涓2^0=1
绛旓細1锛塯.e. = lim(x鈫抜nf.)[(1+1/x)^x]^(1/2)= e^(1/2)锛2锛塯.e. = lim(x鈫抜nf.){{[1-4/(x-1)]^[(x-1)/4]}^8}*[1-4/(x-1)]^3 = [e^(-1)]^8 =e^(-8)锛16锛塯.e. = lim(x鈫0)ln(sinx/x)= ln1 = 0銆
绛旓細鍘熷紡=lim(x鈫0) 2鈭玔0,x] sintdt*sinx/[2x*ln(1+x^2)]=lim(x鈫0) 2鈭玔0,x] sintdt*x/[2x*x^2]=lim(x鈫0) 2鈭玔0,x] sintdt/[2x^2]=lim(x鈫0) 2sinx/(4x)=1/2
