请向这题如何求极限。 请问这题如何求极限
\u8bf7\u5411\u8fd9\u9898\u5982\u4f55\u6c42\u6781\u9650\u7528\u6d1b\u5fc5\u8fbe\u6cd5\u5219\u8fdb\u884c\u6781\u9650\u7684\u6c42\u89e3\uff0c\u8be6\u7ec6\u8fc7\u7a0b\u8bf7\u89c1\u56fe\u7247\uff0c\u5e0c\u671b\u5bf9\u4f60\u6709\u5e2e\u52a9
\u5206\u5b50\u5206\u6bcd\u6709\u7406\u5316
\u540c\u65f6\u4e58\u4e0a[\u221a(2x+1)+3][\u221a(x-2)+\u221a2]\u5f97\uff0c
\u6211\u624b\u5934\u6ca1\u6709\u7eb8\uff0c\u4e0b\u9762\u5c31\u628a\u5206\u5b50\u5206\u6bcd\u5355\u72ec\u5217\u51fa\u6765\u89e3\u91ca
\u5206\u5b50=[\uff082x+1\uff09-9][\u221a(x-2)+\u221a2]=2(x-4)[\u221a(x-2)+\u221a2]
\u5206\u6bcd=[(x-2)-2][\u221a(2x+1)+3]=(x-4)[\u221a(2x+1)+3]
\u539f\u5f0f=lim\uff08x->4\uff092(x-4)[\u221a(x-2)+\u221a2]/(x-4)[\u221a(2x+1)+3]
=lim\uff08x->4\uff092[\u221a(x-2)+\u221a2]/[\u221a(2x+1)+3]
=2\u00d72\u221a2/6
=2\u221a2/3
分子是x^(1+...+n)=x^n(n+1)/2，系数是1
分母是x^n*(n+1)/2，系数是n^n*(n+1)/2
∴lim=1/n^(n(n+1)/2)=n^(－n(n+1)/2)
绛旓細鈭磍im=1/n^(n(n+1)/2)=n^(锛峮(n+1)/2)
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绛旓細鍐嶅彇鏋侀檺=1/2.
绛旓細绠鍗曞垎鏋愪竴涓嬶紝绛旀濡傚浘鎵绀
绛旓細缁撹锛鏋侀檺涓篹鈮2.71828鈥︹︺傜鍚堣繖涓変釜鐗瑰緛鐨勬瀬闄愬奸兘鏄痚銆傛墍浠ワ細lim[x-->0](1+x)^锛1/x锛=e 涓婇潰姝ラ鐪佺暐涓棿姝ラ涓嶅锛屽綋鐒朵笉鐪佷細鏇村ソ鐞嗚В锛屽簲濡備笅锛歭im[x-->鈭瀅[(1-1/x)^x/(1+1/x)^x]=lim[x-->鈭瀅(1-1/x)^x/lim[x-->鈭瀅(1+1/x)^x =lim[x-->鈭瀅{[1+(...
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