\u591a\u5173\u6ce8\u5c40\u90e8\u4e0e\u6574\u4f53\u7684\u6570\u503c\u53d8\u5316\u7684\u4f9d\u8d56\u5173\u7cfb\uff0c\u6293\u4f4f\u4e3b\u8981\u77db\u76fe.<\/p>\n
\u4f8b\uff1a<\/p>\n
\u901a\u9879\u6709\uff1a<\/p>\n k<\/span> \u4ece 1<\/span> \u5230 n<\/span>\uff0c\u901a\u9879\u7684\u6781\u9650\u5bf9\u548c\u5f0f\u7684\u8d21\u732e\u5b8c\u5168\u4e0d\u540c\uff1a<\/p>\n \u5982 k=1,2,3,\\cdots<\/span>\uff0c\u5f53 n \\to \\infty<\/span>\uff0c\\left(\\frac{k}{n}\\right)^n \\to 0<\/span>.<\/p>\n \u6b64\u65f6\u8fd9\u4e9b\u9879\u51e0\u4e4e\u4e3a 0\uff0c\u5bf9\u548c\u5f0f\u51e0\u4e4e\u6ca1\u6709\u8d21\u732e.<\/p>\n \u5982\uff1a<\/p>\n \u6545\u53ea\u6709\u5f53 k<\/span> \u975e\u5e38\u9760\u8fd1 n<\/span> \u7684\u5be5\u5be5\u51e0\u9879\u624d\u5bf9\u548c\u5f0f\u6709\u8d21\u732e\uff0c\u524d\u9762\u4e00\u5927\u5806\u5168\u90e8\u8d8b\u4e8e 0\uff0c\u53ef\u4ee5\u5ffd\u7565.<\/p>\n Guess:<\/p>\n \u8fd9\u91cc\u6211\u4eec\u5c06\u548c\u5f0f\u5206\u6bb5\u4f30\u8ba1<\/strong>\uff0c\u62c6\u6210\u6709\u8d21\u732e\u7684\u90e8\u5206+\u53ef\u4ee5\u5ffd\u7565\u7684\u90e8\u5206<\/strong>\uff0c\u6293\u4f4f\u4e3b\u8981\u77db\u76fe\u3002<\/p>\n \u5f53 0 \\le i \\le n-1<\/span> \u65f6\uff0c\u6709\uff1a<\/p>\n \u56e0\u6b64\uff1a<\/p>\n \u5bf9\u4e8e\u4efb\u610f\u6b63\u6574\u6570 N<\/span>\uff0c\u5f53 n>N<\/span> \u65f6\uff1a<\/p>\n \u4ee4 n \\to \\infty<\/span> \u4e14 N \\to \\infty<\/span>\uff0c\u6709\uff1a<\/p>\n \u7531 (1) (2) \u53ca\u5939\u903c\u5b9a\u7406\u5f97\uff1a<\/p>\n \u8bbe \\displaystyle I_n = n \\int_1^a \\frac{dx}{1+x^n}<\/span>\uff0ca>1<\/span>. \u6c42 \\displaystyle \\lim_{n \\to \\infty} I_n<\/span>.<\/p>\n \u5206\u6bcd\u6709 x^n<\/span>\uff0c\u5148\u6362\u5143 t = x^n<\/span>\uff0c\u6709\uff1a<\/p>\n \u5206\u6790<\/strong>\uff1at<\/span> \u4ece 1<\/span> \u5230 a^n<\/span>\uff0c\u5f53 t<\/span> \u5f88\u5c0f\u65f6\uff0c\u5206\u5b50 \\sqrt[n]{t} \\to 1<\/span>\uff1b\u5f53 t<\/span> \u9760\u8fd1 a^n<\/span> \u65f6\uff0c\u5206\u5b50 \\sqrt[n]{t} \\le a<\/span>\uff0c\u800c\u5206\u6bcd t(1+t)<\/span> \u975e\u5e38\u5927\uff0c\u6574\u4f53\u8d8b\u4e8e 0. \u56e0\u6b64\u79ef\u5206\u503c\u51e0\u4e4e\u53ea\u7531\u524d\u9762\u4e00\u6bb5\u8d21\u732e\uff0c\u6211\u4eec\u540c\u6837\u53ef\u4ee5\u5c1d\u8bd5\u5206\u6bb5\u4f30\u8ba1<\/strong>.<\/p>\n A<\/span> \u662f\u4e00\u4e2a\u5145\u5206\u5927\u7684\u56fa\u5b9a\u5e38\u6570.<\/p>\n \u4ee4 n \\to \\infty<\/span>\uff0c\u518d\u4ee4 A \\to \\infty<\/span>\uff1a \u6545 \\displaystyle \\lim_{n \\to \\infty} I_n = \\ln 2<\/span><\/p>\n \u5f53\u8868\u8fbe\u5f0f\u4e0d\u6613\u8ba1\u7b97\u65f6\uff0c\u53ef\u8003\u8651\u4f7f\u7528\u5939\u6324\u5b9a\u7406. \u57fa\u672c\u60f3\u6cd5\uff1a\u7a81\u51fa\u4e3b\u8981\uff0c\u5ffd\u7565\u6b21\u8981<\/strong>.<\/p>\n \u5206\u6790<\/strong>\uff1a\u5f0f\u4e2d\u51fa\u73b0\u6781\u9650\u3001\u6c42\u548c\u4ee5\u53ca \\frac{k}{n}<\/span>\uff0c\u6709\u70b9\u50cf Riemann \u548c\uff0c\u7136\u800c\u8868\u8fbe\u5f0f\u4e2d\u51fa\u73b0 k^{-1}<\/span>\uff0c\u6211\u4eec\u60f3\u529e\u6cd5\u628a k^{-1}<\/span> \u9879\u653e\u7f29\u6389.<\/p>\n \u5229\u7528 0 < k^{-1} \\le 1<\/span>\uff0c\u6709\uff1a<\/p>\n \u6545\u7531\u5939\u6324\u5b9a\u7406\uff0c\u539f\u6781\u9650\u7684\u503c\u4e3a \\displaystyle \\frac{a-1}{\\ln a}<\/span>.<\/p>\n \u5206\u6790<\/strong>\uff1a\u88ab\u79ef\u5f0f\u8f83\u4e3a\u590d\u6742\uff0c\u76f4\u63a5\u79ef\u5206\u79ef\u4e0d\u51fa\u6765. \u5c1d\u8bd5\u4f30\u8ba1\u5b83\uff0c\u8fd9\u91cc\u6211\u4eec\u6709\u4e24\u79cd\u601d\u8def\uff0c\u4e00\u79cd\u662f\u7528\u79ef\u5206\u4e2d\u503c\u5b9a\u7406\uff0c\u4e00\u79cd\u662f\u76f4\u63a5\u653e\u7f29\u88ab\u79ef\u51fd\u6570.<\/p>\n \u5b58\u5728 \\xi \\in [x, x+1]<\/span>\uff0c\u4f7f\u5f97\uff1a<\/p>\n \u5f53 x\\to+\\infty<\/span>\uff0c\u7531\u4e8e x \\le \\xi \\le x+1<\/span>\uff0c\u6709 \\displaystyle 1 \\le \\frac{\\xi}{x} \\le 1+\\frac{1}{x}<\/span>\uff0c\u5939\u6324\u5f97 \\xi \\sim x \\quad (x \\to +\\infty)<\/span>\uff0c\u6545\uff1a<\/p>\n \u4ee3\u56de\u539f\u6781\u9650\uff1a<\/p>\n \u6ce8\u610f\u5230\uff08\u5176\u5b9e\u6ce8\u610f\u4e0d\u5230\uff09\uff1a<\/p>\n \uff08\u5229\u7528 \\sin t<\/span> \u548c \\cos t<\/span> \u7684\u6709\u754c\u6027\u4ee5\u53ca t \\ge x<\/span>\uff09<\/p>\n \u4e8e\u662f\uff1a<\/p>\n \u4ece\u800cL=0<\/span>.<\/p>\n \u5206\u6790<\/strong>\uff1a\u5206\u5b50\u6709\u4e24\u4e2a\u51fd\u6570\u76f8\u52a0\u51cf\uff0c\u65e0\u6cd5\u7b49\u4ef7\u65e0\u7a77\u5c0f\u66ff\u6362\uff0c\u8003\u8651\u7528\u6cf0\u52d2\u5c55\u5f00. \u8fd9\u91cc\u5bf9\u4e8e (1+x)^{\\frac{2}{x}}<\/span> \u6211\u4eec\u5c06\u5e95\u6570\u653e\u5230\u6307\u6570\u4e0a\u53bb\uff0c\u65b9\u4fbf\u6211\u4eec\u5c55\u5f00.<\/p>\n \u4e8e\u662f\uff1a<\/p>\n \u6211\u4eec\u5bf9 (\\cos x)^{\\sin x}<\/span> \u91c7\u7528\u540c\u4e0a\u7684\u5904\u7406\uff1a<\/p>\n \u7136\u540e\u4f7f\u7528\u7b49\u4ef7\u65e0\u7a77\u5c0f\uff1a<\/p>\n \u5c06\u6839\u5f0f\u5199\u6210\u6307\u6570\u5f62\u5f0f\uff0c\u5e76\u63d0\u53d6\u516c\u56e0\u5f0f\uff1a<\/p>\n \u8fd9\u6837\u5904\u7406\u540e\u6211\u4eec\u80fd\u51d1\u51fa a^x-1<\/span> \u7684\u7b49\u4ef7\u65e0\u7a77\u5c0f\u5f62\u5f0f\uff1a<\/p>\n \u4e8e\u662f\u539f\u6781\u9650\uff1a<\/p>\n \u7c7b\u4f3c\u5730\uff0c\u4e0b\u9898\u4e5f\u91c7\u7528\u540c\u6837\u7684\u5904\u7406\u601d\u8def\uff0c\u5c06\u5206\u5b50\u63d0\u53d6\u4e00\u4e2a x<\/span> \u540e\u7b49\u4ef7\u65e0\u7a77\u5c0f\u5904\u7406\uff1a<\/p>\n Stolz \u5b9a\u7406<\/strong> \u8bbe\u6570\u5217 \\{x_n\\}<\/span>\uff0c\\{y_n\\}<\/span> \u6ee1\u8db3\u4e0b\u5217\u6761\u4ef6\u4e4b\u4e00\uff1a<\/p>\n \u82e5 Stolz \u5b9a\u7406\u53ef\u4ee5\u770b\u4f5c\u662f\u79bb\u6563\u60c5\u51b5\u4e0b\u7684\u6d1b\u5fc5\u8fbe\u6cd5\u5219.<\/p>\n \u8bbe 0 < x_1 < 1<\/span>\uff0cx_{n+1} = x_n (1-x_n)<\/span>\uff0cn=1,2,\\cdots<\/span>. \u89e3<\/strong> \uff081\uff09\u7531\u6570\u5b66\u5f52\u7eb3\u6cd5\u6613\u8bc1 0 < x_n < 1<\/span>\uff0c \uff082\uff09\u7531 Stolz \u5b9a\u7406\uff1a \u8bbe\u6570\u5217 \\{x_n\\}_{n=1}^{\\infty}<\/span> \u6ee1\u8db3 \\lim\\limits_{n\\to\\infty}(x_n-x_{n-2})=0<\/span>\uff0c\u6c42\u8bc1\uff1a \u8bc1<\/strong> \u7531 Stolz \u5b9a\u7406\uff1a \u5206\u6790<\/strong>\uff1a\u9898\u8bbe\u6761\u4ef6\u6709 x_n-x_{n-2}<\/span>\uff0c\u5f85\u8bc1\u6781\u9650\u4e2d\u662f x_n-x_{n-1}<\/span>\uff0c\u6ce8\u610f\u5230 x_n-x_{n-2} = (x_n-x_{n-1}) + (x_{n-1}-x_{n-2})<\/span>\uff0c\u7b49\u5f0f\u53f3\u8fb9\u521a\u597d\u662f\u8fde\u7eed\u4e24\u9879\uff0c\u60f3\u5230\u53ef\u4ee5\u7528 Stolz \u5b9a\u7406\uff0c\u4f46\u662f\u5b9a\u7406\u4e2d\u662f\u4e24\u9879\u4e4b\u5dee\uff0c\u8fd9\u91cc\u5374\u662f\u4e24\u9879\u4e4b\u548c\uff0c\u4e8e\u662f\u6709\u795e\u4e4b\u4e00\u624b<\/strong>\u2014\u2014\u6784\u9020 (-1)^n(x_n-x_{n-1})<\/span>.<\/p>\n \u5206\u6790<\/strong>\uff1a\u60f3\u8981\u8f6c\u5316\u4e3a Riemann \u548c\u5f62\u5f0f\uff0c\u4f46\u662f\u6709\u5e38\u6570 1<\/span>\uff0c\u60f3\u529e\u6cd5\u628a 1<\/span> \u8f6c\u5316\u4e3a\u6c42\u548c\u5f62\u5f0f\u2014\u2014 \\displaystyle 1 = \\sum_{k=1}^n \\frac{1}{n}<\/span>.<\/p>\n \u5176\u4e2d \\dfrac{\\sqrt{k}}{n}\\to0<\/span>\uff08\u6216\u7528 0 < k \\le n<\/span> \u5939\u6324\uff09.<\/p>\n \u8bbe f(x)<\/span> \u5728\u95ed\u533a\u95f4 [0,1]<\/span> \u4e0a\u6709\u8fde\u7eed\u5bfc\u6570\uff0cf(0)=0<\/span>\uff0cf(1)=1<\/span>. \u5206\u6790<\/strong>\uff1a\u8fd9\u9053\u9898\u540c\u6837\u60f3\u8981\u8f6c\u5316\u4e3a Riemann \u548c\uff0c\u4f46\u662f\u5f0f\u4e2d\u51fa\u73b0\u79ef\u5206\uff0c\u540c\u6837\u7684\u601d\u8def\u5c06\u79ef\u5206\u8f6c\u5316\u4e3a\u6c42\u548c. \u7136\u540e\u540e\u9762\u4e00\u987f\u884c\u4e91\u6d41\u6c34...\u6211\u771f\u7684\u60f3\u4e0d\u51fa\u6765...\u89c1\u4e0b.<\/p>\n \u8fd8\u6709\u4e24\u4e2a\u795e\u79d8\u9898\u76ee\u592a\u795e\u79d8\u4e86\uff0c\u540e\u9762\u518d\u5355\u72ec\u770b\u770b. \u4ee5\u53ca\u4e0e\u7ea7\u6570\u76f8\u5173\u7684\u9898\uff0c\u540e\u9762\u4e5f\u5355\u72ec\u53bb\u7814\u7a76\u5427.<\/p>\n","protected":false},"excerpt":{"rendered":" \u795e\u79d8\u534e\u79d1 CMC \u6821\u8bad\u7b2c\u4e00\u671f\u2014\u2014\u6781\u9650……\u611f\u89c9\u6211\u662ffw<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"emotion":"","emotion_color":"","title_style":"","license":"","footnotes":""},"categories":[10],"tags":[29,31,30],"class_list":["post-268","post","type-post","status-publish","format-standard","hentry","category-math","tag-cmc","tag-31","tag-30"],"_links":{"self":[{"href":"https:\/\/blog.huajimc.cn\/index.php\/wp-json\/wp\/v2\/posts\/268","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.huajimc.cn\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.huajimc.cn\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.huajimc.cn\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.huajimc.cn\/index.php\/wp-json\/wp\/v2\/comments?post=268"}],"version-history":[{"count":52,"href":"https:\/\/blog.huajimc.cn\/index.php\/wp-json\/wp\/v2\/posts\/268\/revisions"}],"predecessor-version":[{"id":320,"href":"https:\/\/blog.huajimc.cn\/index.php\/wp-json\/wp\/v2\/posts\/268\/revisions\/320"}],"wp:attachment":[{"href":"https:\/\/blog.huajimc.cn\/index.php\/wp-json\/wp\/v2\/media?parent=268"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.huajimc.cn\/index.php\/wp-json\/wp\/v2\/categories?post=268"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.huajimc.cn\/index.php\/wp-json\/wp\/v2\/tags?post=268"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\frac{k^n}{n^n}
\n<\/span><\/p>\n
\n\\frac{k^n}{n^n}=\\left(\\frac{k}{n}\\right)^n
\n<\/span><\/p>\n1. k \\ll n<\/span><\/h4>\n
2. k<\/span> \u9760\u8fd1 n<\/span> \u65f6\uff0c\u5373 k=n-i<\/span><\/h4>\n
\n\\left(\\frac{k}{n}\\right)^n=\\left(\\frac{n-i}{n}\\right)^n=\\left(1-\\frac{i}{n}\\right)^n \\to e^{-i}
\n<\/span><\/p>\n\n
\n\\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\frac{k^n}{n^n} = 1 + e^{-1} + e^{-2} + \\cdots = \\frac{e}{e-1}
\n<\/span><\/p>\n\u4e25\u683c\u8bc1\u660e<\/h3>\n
\u4e0a\u754c<\/h4>\n
\n\\left(1-\\frac{i}{n}\\right)^n \\le e^{-i}
\n<\/span><\/p>\n
\nS_n = \\sum_{i=0}^{n-1} \\left(1-\\frac{i}{n}\\right)^n \\le \\sum_{i=0}^{n-1} e^{-i} < \\sum_{i=0}^{\\infty} e^{-i}
\n\\tag{1}
\n<\/span><\/p>\n\u4e0b\u754c<\/h4>\n
\nS_n = \\sum_{i=0}^{n-1} \\left(1-\\frac{i}{n}\\right)^n \\ge \\sum_{i=0}^{N} e^{-i}
\n<\/span><\/p>\n
\n\\begin{equation}
\n\\lim_{n \\to \\infty} S_n \\ge \\sum_{i=0}^{\\infty} e^{-i}
\n\\tag{2}
\n\\end{equation}
\n<\/span><\/p>\n
\n\\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\frac{k^n}{n^n} = \\sum_{i=0}^{\\infty} e^{-i} = \\frac{e}{e-1}
\n<\/span><\/p>\n\u4e00\u9053\u7c7b\u4f3c\u7684\u9898<\/h3>\n
\nI_n = \\int_1^{a^n} \\frac{\\sqrt[n]{t}}{t(1+t)} \\, dt
\n<\/span><\/p>\n
\nI_n = \\int_1^A \\frac{\\sqrt[n]{t}}{t(1+t)} \\, dt + \\int_A^{a^n} \\frac{\\sqrt[n]{t}}{t(1+t)} \\, dt
\n<\/span><\/p>\n
\n
\n\\int_1^A \\frac{\\sqrt[n]{t}}{t(1+t)} \\, dt \\to \\int_1^A \\frac{1}{t(1+t)} \\, dt \\to \\int_1^{+\\infty} \\frac{1}{t(1+t)} \\, dt = \\ln 2
\n<\/span><\/p>\n
\n0 < \\int_A^{a^n} \\frac{\\sqrt[n]{t}}{t(1+t)} \\, dt \\le a \\int_A^{a^n} \\frac{dt}{t(1+t)} = a \\ln{\\left(1+\\frac{1}{A}\\right)} < \\frac{a}{A} \\to 0
\n<\/span><\/p>\n\u4e8c\u3001\u5939\u6324\u5b9a\u7406<\/h2>\n
\u4f8b 1<\/h3>\n
\n\\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\frac{a^{\\frac{k}{n}}}{n + (a-1) k^{-1}} \\qquad (a > 1)
\n<\/span><\/p>\n
\n\\sum_{k=1}^{n} \\frac{a^{\\frac{k}{n}}}{n + (a-1)} \\le \\sum_{k=1}^{n} \\frac{a^{\\frac{k}{n}}}{n + (a-1) k^{-1}} < \\sum_{k=1}^{n} \\frac{a^{\\frac{k}{n}}}{n}
\n<\/span><\/p>\n
\n\\lim_{n\\to\\infty}\\sum_{k=1}^{n}\\frac{a^{\\frac{k}{n}}}{n}
\n=\\int_{0}^{1}a^{x}dx
\n=\\frac{a-1}{\\ln a}
\n<\/span><\/p>\n
\n\\lim_{n\\to\\infty}\\sum_{k=1}^{n}\\frac{a^{\\frac{k}{n}}}{n+(a-1)}
\n=\\lim_{n\\to\\infty}\\frac{n}{n+a-1}\\lim_{n\\to\\infty}\\sum_{k=1}^{n}\\frac{a^{\\frac{k}{n}}}{n}
\n=\\frac{a-1}{\\ln a}
\n<\/span><\/p>\n\u4f8b 2<\/h3>\n
\n\\lim_{x\\to+\\infty}\\sqrt[3]{x}\\int_{x}^{x+1}\\frac{\\sin t}{\\sqrt{t+\\cos t}}\\,dt
\n<\/span><\/p>\n\u6cd5\u4e00\uff1a\u79ef\u5206\u4e2d\u503c\u5b9a\u7406<\/h4>\n
\n\\int_{x}^{x+1}\\frac{\\sin t}{\\sqrt{t+\\cos t}}\\,dt = \\frac{\\sin \\xi}{\\sqrt{\\xi+\\cos \\xi}}
\n<\/span><\/p>\n
\n\\frac{\\sin \\xi}{\\sqrt{\\xi+\\cos \\xi}} \\sim \\frac{\\sin \\xi}{\\sqrt{\\xi}} \\sim \\frac{\\sin \\xi}{\\sqrt{x}}
\n<\/span><\/p>\n
\nL = \\lim_{x\\to+\\infty}\\sqrt[3]{x}\\cdot\\frac{\\sin \\xi}{\\sqrt{x}} = \\lim_{x\\to+\\infty} x^{-\\frac{1}{6}} \\cdot \\sin \\xi \\xlongequal{\\text{\u65e0\u7a77\u5c0f\u00d7\u6709\u754c}} 0
\n<\/span><\/p>\n\u6cd5\u4e8c\uff1a\u653e\u7f29<\/h4>\n
\n\\left|\\int_{x}^{x+1}\\frac{\\sin t}{\\sqrt{t+\\cos t}}\\,dt\\right| \\le \\int_{x}^{x+1}\\left|\\frac{\\sin t}{\\sqrt{t+\\cos t}}\\right|\\,dt \\le \\int_{x}^{x+1}\\frac{1}{\\sqrt{t-1}}\\,dt \\le \\int_{x}^{x+1}\\frac{1}{\\sqrt{x-1}}\\,dt = \\frac{1}{\\sqrt{x-1}}
\n<\/span><\/p>\n
\n|L| \\le \\lim_{x\\to+\\infty}\\sqrt[3]{x}\\cdot\\frac{1}{\\sqrt{x-1}} = 0
\n<\/span><\/p>\n\u4e09\u3001Taylor \u5c55\u5f00<\/h2>\n
\u4f8b 1<\/h3>\n
\n\\lim_{x\\to 0}\\frac{(1+x)^{\\frac{2}{x}}-e^{2}\\big(1-\\ln(1+x)\\big)}{x}
\n<\/span><\/p>\n
\n\\begin{aligned}
\n&\\ln (1+x) = x-\\frac12 x^2+o(x^2)
\n\\\\
\n&\\textcolor{red}{(1+x)^{\\frac{2}{x}} = e^{\\frac{2}{x}\\ln(1+x)} = e^{2}\\cdot e^{-x+o(x)} = e^{2}\\big(1-x+o(x)\\big) \\qquad (\\star)}
\n\\\\
\n&e^{2}\\big(1-\\ln(1+x)\\big) = e^{2}\\left(1-x+\\frac12 x^2+o(x^2)\\right)
\n\\end{aligned}
\n<\/span><\/p>\n
\n\\begin{aligned}
\n&\\lim_{x\\to 0}\\frac{(1+x)^{\\frac{2}{x}}-e^{2}\\big(1-\\ln(1+x)\\big)}{x} \\\\
\n=&\\lim_{x\\to 0}\\frac{e^{2}\\big(1-x+o(x)\\big)-e^{2}\\left(1-x+\\frac12 x^2+o(x^2)\\right)}{x} \\\\
\n=&0
\n\\end{aligned}
\n<\/span><\/p>\n\u4f8b 2<\/h3>\n
\n\\lim_{x\\to 0}\\frac{1-(\\cos x)^{\\sin x}}{x^3}
\n<\/span><\/p>\n
\n(\\cos x)^{\\sin x}=e^{\\sin x \\ln \\cos x}=1+\\sin x \\ln \\cos x+o(\\sin x \\ln \\cos x)
\n<\/span><\/p>\n
\n\\sin x \\ln \\cos x=\\sin x \\ln(\\cos x-1+1)\\sim x(\\cos x-1)\\sim -\\frac12 x^3
\n<\/span><\/p>\n\u4f8b 3<\/h3>\n
\n\\lim_{n\\to+\\infty} n^2 (\\sqrt[n]{2026}-\\sqrt[n+1]{2026})
\n<\/span><\/p>\n
\n\\sqrt[n]{2026}-\\sqrt[n+1]{2026} = 2026^{\\frac{1}{n}} - 2026^{\\frac{1}{n+1}} = 2026^{\\frac{1}{n+1}} (2026^{\\frac{1}{n} - \\frac{1}{n+1}} - 1)
\n<\/span><\/p>\n
\n2026^{\\frac{1}{n} - \\frac{1}{n+1}} - 1 \\sim \\frac{\\ln 2026}{n(n+1)}
\n<\/span><\/p>\n
\nL = \\lim_{n\\to+\\infty} n^2 \\cdot \\frac{\\ln 2026}{n(n+1)} = \\ln 2026
\n<\/span><\/p>\n
\n\\lim_{x\\to 0}\\frac{\\sqrt[3]{\\sin x^3}-x}{x^7}
\n<\/span><\/p>\n\u56db\u3001Stolz \u5b9a\u7406<\/h2>\n
\n
\n
\n\\lim_{n\\to\\infty} \\frac{y_{n+1}-y_n}{x_{n+1}-x_n} = a
\n<\/span>
\n\u5219
\n
\n\\lim_{n\\to\\infty} \\frac{y_n}{x_n} = \\lim_{n\\to\\infty} \\frac{y_{n+1}-y_n}{x_{n+1}-x_n} = a.
\n<\/span><\/p>\n\u4f8b 1<\/h3>\n
\n\u8bc1\u660e\uff1a\uff081\uff09\\displaystyle \\lim_{n\\to\\infty} x_n=0<\/span>\uff1b\uff082\uff09\\displaystyle \\lim_{n\\to\\infty} nx_n=1<\/span>.<\/p>\n
\n\u4e8e\u662f x_{n+1} = x_n (1-x_n) < x_n<\/span>\uff0c\u4ece\u800c \\displaystyle \\lim_{n\\to\\infty} x_n = l<\/span> \u5b58\u5728.
\n\u7531 \\displaystyle \\lim_{n\\to\\infty} x_{n+1} = \\lim_{n\\to\\infty} x_n (1-x_n)<\/span>\uff0c
\n\u6709 l=l(1-l)<\/span>\uff0c\u7b80\u5355\u8ba8\u8bba\u53ef\u77e5 l=0<\/span>.<\/p>\n
\n
\n\\begin{aligned}
\n\\lim_{n\\to\\infty} nx_n &= \\lim_{n\\to\\infty} \\frac{n}{\\dfrac{1}{x_n}} = \\lim_{n\\to\\infty} \\frac{n-(n-1)}{\\dfrac{1}{x_n} - \\dfrac{1}{x_{n-1}}} \\\\
\n&= \\lim_{n\\to\\infty} \\frac{x_n x_{n-1}}{x_{n-1}-x_n} = \\lim_{n\\to\\infty} \\frac{x_n}{x_{n-1}} \\\\
\n&= \\lim_{n\\to\\infty} (1-x_{n-1}) = 1
\n\\end{aligned}
\n<\/span><\/p>\n\u4f8b 2<\/h3>\n
\n
\n\\lim_{n\\to\\infty}\\frac{x_n-x_{n-1}}{n}=0
\n<\/span><\/p>\n
\n
\n\\begin{aligned}
\n&\\lim_{n\\to\\infty}\\frac{(-1)^n(x_n-x_{n-1})}{n} \\\\
\n=& \\lim_{n\\to\\infty}\\frac{(-1)^n(x_n-x_{n-1})-(-1)^{n-1}(x_{n-1}-x_{n-2})}{n-(n-1)} \\\\
\n=& \\lim_{n\\to\\infty}(-1)^n(x_n-x_{n-2}) = 0
\n\\end{aligned}
\n<\/span><\/p>\n\u4e94\u3001\u4e0e\u79ef\u5206\u76f8\u5173\u7684\u6781\u9650<\/h2>\n
\n
\u4f8b 1<\/h3>\n
\n\\lim_{n\\to\\infty} \\sqrt{n} (1 - \\sum_{k=1}^n \\frac{1}{n+\\sqrt{k}})
\n<\/span><\/p>\n
\n\\sqrt{n} (1 - \\sum_{k=1}^n \\frac{1}{n+\\sqrt{k}}) = \\sqrt{n} \\sum_{k=1}^n \\frac{\\sqrt{k}}{n(n+\\sqrt{k})} = \\sum_{k=1}^n \\frac{\\sqrt{\\dfrac{k}{n}}}{1+\\dfrac{\\sqrt{k}}{n}}\\cdot\\frac1n
\n<\/span><\/p>\n\u4f8b 2<\/h3>\n
\n\\lim_{n\\to\\infty}\\frac{1}{\\sqrt{n}}\\left(\\frac{n+1}{2}-\\sum_{k=1}^n \\frac{k}{n+\\sqrt{k}}\\right)
\n<\/span><\/p>\n
\n\\begin{aligned}
\n&\\frac{1}{\\sqrt{n}}\\left(\\frac{n+1}{2}-\\sum_{k=1}^n \\frac{k}{n+\\sqrt{k}}\\right)
\n=\\frac{1}{\\sqrt{n}}\\left(\\frac{n(n+1)}{2}\\cdot\\frac1n-\\sum_{k=1}^n \\frac{k}{n+\\sqrt{k}}\\right) \\\\
\n=&\\frac{1}{\\sqrt{n}}\\left(\\sum_{k=1}^n \\frac{k}{n}-\\sum_{k=1}^n \\frac{k}{n+\\sqrt{k}}\\right)
\n=\\frac{1}{\\sqrt{n}}\\sum_{k=1}^n \\frac{k\\sqrt{k}}{n(n+\\sqrt{k})}
\n=\\sum_{k=1}^n \\frac{\\dfrac{k}{n}\\sqrt{\\dfrac{k}{n}}}{1+\\dfrac{\\sqrt{k}}{n}}\\cdot \\frac1n
\n\\end{aligned}
\n<\/span><\/p>\n\u4f8b 3<\/h3>\n
\n\u8bc1\u660e\uff1a
\n
\n\\lim_{n\\to\\infty} n\\left(\\int_{0}^{1} f(x)dx -\\frac1n\\sum_{k=1}^{n} f\\left(\\frac{k}{n}\\right)\\right)=-\\frac12.
\n<\/span><\/p>\n
\n\\begin{align*}
\n&\\ n\\left(\\int_{0}^{1} f(x)\\,dx - \\frac1n\\sum_{k=1}^{n} f\\left(\\frac{k}{n}\\right)\\right) \\\\
\n&= n\\left(\\sum_{k=1}^{n}\\int_{\\frac{k-1}{n}}^{\\frac{k}{n}} f(x)\\,dx - \\frac1n\\sum_{k=1}^{n} f\\left(\\frac{k}{n}\\right)\\right) \\tag{\u5c06\u79ef\u5206\u8f6c\u5316\u4e3a\u6c42\u548c} \\\\
\n&= n\\sum_{k=1}^{n}\\int_{\\frac{k-1}{n}}^{\\frac{k}{n}} \\left(f(x) - f\\left(\\frac{k}{n}\\right)\\right) dx \\tag{\u5408\u5e76\u6c42\u548c\u9879} \\\\
\n&= -n\\sum_{k=1}^{n}\\int_{\\frac{k-1}{n}}^{\\frac{k}{n}} \\left(x - \\frac{k-1}{n}\\right) f'(x)\\,dx \\tag{\u5206\u90e8\u79ef\u5206} \\\\
\n&= -n\\sum_{k=1}^{n} f'(\\xi_k)\\int_{\\frac{k-1}{n}}^{\\frac{k}{n}} \\left(x - \\frac{k-1}{n}\\right) dx \\tag{\u79ef\u5206\u4e2d\u503c\u5b9a\u7406} \\\\
\n&= -\\frac12\\sum_{k=1}^{n} f'(\\xi_k)\\frac{1}{n} \\to -\\frac12\\int_{0}^{1} f'(x)\\,dx = -\\frac12 \\tag{Riemann \u548c}
\n\\end{align*}
\n<\/span><\/p>\n
\n