lim1/x在x→0是等于0吗？为什么。0-时为-∞，0+时为+∞，也可以说它有极限？ 求极限lim e^(1/x)=0 x→0-极限怎么算来的？
\u4e3a\u4ec0\u4e48lim \uff08x\u8d8b\u4e8e0\uff09(1+x)^(1/x)\u7b49\u4e8ee\uff1f\u56e0\u4e3ax\u8d8b\u4e8e0\uff0c\u6240\u4ee5lim[(1+x)^(1/x)]=lim(1+x)^\u221e=e
\u89e3\u9898\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u539f\u5f0f = lim (e^(ln(1+x)/x) -e)/x
=lim e(e^(ln(1+x)/x - 1) -1 ) /x
=lim e(ln(1+x)/x -1)/x
=e lim (ln(1+x)-x)/x²
=e lim (1/(1+x)-1) / 2x
=e lim -x/(2x(1+x))
=lim[(1+x)^(1/x)]
=lim(1+x)^\u221e
=e
\u6269\u5c55\u8d44\u6599\u6c42\u51fd\u6570\u6781\u9650\u7684\u65b9\u6cd5\uff1a
\u5229\u7528\u51fd\u6570\u8fde\u7eed\u6027\uff0c\u76f4\u63a5\u5c06\u8d8b\u5411\u503c\u5e26\u5165\u51fd\u6570\u81ea\u53d8\u91cf\u4e2d\uff0c\u6b64\u65f6\u8981\u8981\u6c42\u5206\u6bcd\u4e0d\u80fd\u4e3a0\u3002
\u5f53\u5206\u6bcd\u7b49\u4e8e\u96f6\u65f6\uff0c\u5c31\u4e0d\u80fd\u5c06\u8d8b\u5411\u503c\u76f4\u63a5\u4ee3\u5165\u5206\u6bcd\uff0c\u56e0\u5f0f\u5206\u89e3\uff0c\u901a\u8fc7\u7ea6\u5206\u4f7f\u5206\u6bcd\u4e0d\u4f1a\u4e3a\u96f6\u3002\u82e5\u5206\u6bcd\u51fa\u73b0\u6839\u53f7\uff0c\u53ef\u4ee5\u914d\u4e00\u4e2a\u56e0\u5b50\u4f7f\u6839\u53f7\u53bb\u9664\u3002
\u5982\u679c\u8d8b\u5411\u4e8e\u65e0\u7a77\uff0c\u5206\u5b50\u5206\u6bcd\u53ef\u4ee5\u540c\u65f6\u9664\u4ee5\u81ea\u53d8\u91cf\u7684\u6700\u9ad8\u6b21\u65b9\u3002\uff08\u901a\u5e38\u4f1a\u7528\u5230\u8fd9\u4e2a\u5b9a\u7406\uff1a\u65e0\u7a77\u5927\u7684\u5012\u6570\u4e3a\u65e0\u7a77\u5c0f\uff09\u3002
\u91c7\u7528\u6d1b\u5fc5\u8fbe\u6cd5\u5219\u6c42\u6781\u9650\uff0c\u5f53\u9047\u5230\u5206\u5f0f0/0\u6216\u8005\u221e/\u221e\u65f6\u53ef\u4ee5\u91c7\u7528\u6d1b\u5fc5\u8fbe\uff0c\u5176\u4ed6\u5f62\u5f0f\u4e5f\u53ef\u4ee5\u901a\u8fc7\u53d8\u6362\u6210\u6b64\u5f62\u5f0f\u3002\u7b26\u5408\u5f62\u5f0f\u7684\u5206\u5f0f\u7684\u6781\u9650\u7b49\u4e8e\u5206\u5f0f\u7684\u5206\u5b50\u5206\u6bcd\u540c\u65f6\u6c42\u5bfc\u3002
\u7531\u4e8ef(x) = e^(1/x)-1\u5728x=1\u5904\u8fde\u7eed,\u6545\u6709\u8fde\u7eed\u51fd\u6570\u5b9a\u4e49\u77e5\u9053\uff1af(x)\u5728x=1\u5904\u7684\u6781\u9650\u5c31\u662ff(1),\u8ba1\u7b97\u53ef\u5f97f(x) = 0\u3002
\u5982\u679cf(x) = e^(1/(1-x)),\u90a3\u4e48x-->1\u65f6,\u5de6\u6781\u9650\u4e3a0,\u53f3\u6781\u9650\u4e3a\u6b63\u65e0\u7a77\u3002
\u5176\u5b9e\u5f53x\u8d8b\u4e8e1\u65f6,1/(1-x)\u662f\u8d8b\u4e8e\u65e0\u7a77\u7684(x1\u65f6\u8d8b\u4e8e\u6b63\u65e0\u7a77),\u4ece\u800ce^(1/(1-x))\u6709\u4e24\u79cd\u6781\u9650\u3002
\u62d3\u5c55\u8d44\u6599\uff1a
\u9ad8\u7b49\u6570\u5b66\u6c42\u6781\u9650,\u6c42lim[1/e*(1+x)^\uff081/x\uff09]^(1/x) \u3010x\u8d8b\u4e8e0\u3011
\u5982\u9898\uff1a\u6c42lim[\uff081/e\uff09*(1+x)^\uff081/x\uff09]^(1/x) \u3010x\u8d8b\u4e8e0\u3011
\u89e3\u7b54\uff1a
lim[\uff081/e\uff09*(1+x)^\uff081/x\uff09]^(1/x)
=lim[1+\uff08(1+x)^\uff081/x\uff09-e)/e]^[[e/\uff08(1+x)^\uff081/x\uff09-e)]*[\uff08(1+x)^\uff081/x\uff09-e)/ex ]]
=lime^\uff08(1+x)^\uff081/x\uff09-e)/ex
lim((1+x)^\uff081/x\uff09-e)/ex
=lim(x-(1+x)ln(1+x))/x^2
=-1/2
\u6240\u4ee5lim[\uff081/e\uff09*(1+x)^\uff081/x\uff09]^(1/x) \u3010x\u8d8b\u4e8e0\u3011\uff1de^(-1/2)\u3002
绛旓細鎮ㄥソ锛屽父鏁扮殑鏋侀檺涓哄父鏁版湰韬紝鎵浠ワ紝lim0=0, lim1=1 姣斿璇佹槑limc=c(x鈫抶銆)锛屾澶刢涓轰竴甯告暟 璇 杩欓噷涓╢锛坸锛-A涓紳涓╟-c涓紳0锛屽洜姝や换鎰廵psilon锛0锛屽彲鍙杁elta锛0锛屽綋0锛滀辅x-x銆俵锛渄elta鏃讹紝鑳戒娇涓嶇瓑寮忎辅f锛坸锛-A l锛漧c-c涓=0锛渆psilon鎴愮珛銆傛墍浠imc锛漜锛坸鈫抶銆傦級绁濆涔犳剦...
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绛旓細,lim1=1 甯告暟鐨勬瀬闄=鏈韩銆
绛旓細涓嶇瓑lim1鏄棤闄愯秼杩1,骞朵笉绛変簬1锛岃宭ne绛変簬1锛屼袱鑰呬笉绛
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绛旓細iim鍚庨潰鐨勮〃杈惧紡鏄父鏁帮紝閮戒笌鍙樻暟x鐨勫兼棤鍏充簡 鈭limx->0- 1=1
绛旓細涓句緥璇存槑锛lim(x^2+ax+b)/sin(x^2-1)鍒嗘瘝褰搙鈫1鏃舵瀬闄愭槸0,鏁呭垎瀛愭瀬闄愪篃搴斾负0,鍚﹀垯鏈鍚庣粨鏋滀笉鍙兘鏄父鏁.鍗1^2+a*1+b=a+b+1=0,瑙ｅ緱b=-a-1鍘熷紡=lim(x^2+ax+b)/(x^2-1)(绛変环鏃犵┓灏忎唬鎹)=lim(x^2+ax-a-1)/(x^2-1)=lim[(x^2-1)+a(x-1)]/(x^2-1)=1+alim...
绛旓細鍥炵瓟锛氭妸涓婇潰鐨勫紡瀛愬拰涓嬮潰鐨勫紡瀛愬寲绠寰limX涔涒垶X鈥2/X+1 鍥犱负X涔涒垶鎵浠鈥2涔涒垶X+1涔涒垶鍖栫畝=1
绛旓細lim锛屾槸鏋侀檺鏁板鍙枫傛槸涓涓爣璇嗗姛鑳斤紝琛ㄧず鈥滄眰鏋侀檺鈥濄傚叿浣撶殑璇漧im涓嬮潰杩樻湁涓涓+绗﹀彿鈥濓紙瓒嬩簬姝ｆ棤绌凤級锛屸-绗﹀彿鈥濓紙瓒嬩簬璐熸棤绌凤級锛屽叾鍏蜂綋璁＄畻涓句緥濡備笅鍥炬墍绀猴細
