{"id":2129,"date":"2023-04-02T09:55:44","date_gmt":"2023-04-02T01:55:44","guid":{"rendered":"https:\/\/www.appblog.cn\/?p=2129"},"modified":"2023-04-05T20:41:41","modified_gmt":"2023-04-05T12:41:41","slug":"fundamentals-of-high-school-mathematics-limits-and-criteria-of-sequences","status":"publish","type":"post","link":"https:\/\/www.appblog.cn\/index.php\/2023\/04\/02\/fundamentals-of-high-school-mathematics-limits-and-criteria-of-sequences\/","title":{"rendered":"\u9ad8\u4e2d\u6570\u5b66\u57fa\u7840\uff1a\u6570\u5217\u7684\u6781\u9650\u53ca\u5176\u51c6\u5219"},"content":{"rendered":"

\u6570\u5217\u6781\u9650\u7684\u5b9a\u4e49<\/h2>\n

\u8bbe$\\{x_n\\}$\u4e3a\u4e00\u6570\u5217\uff0c\u82e5\u6709\u5e38\u6570$a$\uff0c\u5bf9\u4efb\u610f\u7ed9\u5b9a\u7684\u6b63\u6570$\\varepsilon$\uff08\u65e0\u8bba$\\varepsilon$\u6709\u591a\u5c0f\uff09\uff0c\u603b\u5b58\u5728\u6b63\u6574\u6570$N$\uff0c\u4f7f\u5f53$n>N$\u65f6\uff0c\u4e0d\u7b49\u5f0f$|x_n-a|<\\varepsilon$\u6052\u6210\u7acb\uff0c\u5219\u79f0$a$\u662f\u6570\u5217$\\{x_n\\}$\u7684\u6781\u9650\u6216\u79f0$\\{x_n\\}$\u6536\u655b\u4e8e$a$\uff0c\u8bb0\u4e3a<\/p>\n

<\/p>\n

$$
\n\\lim \\limits_{x \\to \u221e} x_n = a \\,\\, \u6216 \\,\\, x_n \\rightarrow a \\, (n \\rightarrow \u221e)
\n$$<\/p>\n

\u2460 $\\varepsilon$\u662f\u4efb\u610f\u7684\uff0c\u8fd9\u6837\u624d\u80fd\u8868\u793a\u65e0\u9650\u63a5\u8fd1
\n\u2461 $N$\u662f\u76f8\u5e94\u4e8e$\\varepsilon$\u7684\uff0c\u53ea\u8981$N$\u5b58\u5728\uff0c\u800c\u4e0d\u5fc5\u627e\u5176\u6700\u5c0f\u503c<\/p>\n

\u82e5\u8fd9\u6837\u7684$a$\u4e0d\u5b58\u5728\uff0c\u5219\u79f0\u6570\u5217$\\{x_n\\}$\u65e0\u6781\u9650\u6216$\\{xn\\}$\u53d1\u6563$\\lim \\limits<\/em>{x \\to \u221e} x_n$\u4e0d\u5b58\u5728\u3002<\/p>\n

\u4f8b\u4e00<\/h3>\n

\u5df2\u77e5\uff1a<\/p>\n

$$
\nx_n=\\frac{n+(-1)^n}{n}
\n$$<\/p>\n

\u8bc1\u660e\u6570\u5217$\\{x_n\\}$\u7684\u6781\u9650\u4e3a1\u3002<\/p>\n

\u8bc1\uff1a<\/p>\n

$$
\n|x_n-1|=|\\frac{n+(-1)^n}{n}-1|=\\frac{1}{n}
\n$$<\/p>\n

$\\forall \\varepsilon > 0$\uff0c\u6b32\u4f7f$|x_n-1|<\\varepsilon$\uff0c\u5373$\\frac{1}{n}<\\varepsilon$\uff0c\u53ea\u8981$n>\\frac{1}{\\varepsilon}$<\/p>\n

\u56e0\u6b64\uff0c\u53d6$N=[\\frac{1}{\\varepsilon}]$\uff0c\u5f53$n>N$\u65f6\uff0c\u5c31\u6709<\/p>\n

$$
\n|\\frac{n+(-1)^n}{n}-1| < \\varepsilon
\n$$<\/p>\n

\u6545<\/p>\n

$$
\n\\lim \\limits_{x \\to \u221e} xn = \\lim \\limits<\/em>{x \\to \u221e} \\frac{n+(-1)^n}{n} = 1
\n$$<\/p>\n

\u4f8b\u4e8c<\/h3>\n

\u8bc1\u660e\uff1a<\/p>\n

$$
\n\\lim \\limits_{x \\to \u221e} \\frac{\\cos \\frac{n\\pi}{2}}{n} = 0
\n$$<\/p>\n

\u5206\u6790\uff1a<\/p>\n

$$
\n|x_n-0| = \\frac{|\\cos \\frac{n\\pi}{2}|}{n}
\n$$<\/p>\n

$\\forall \\varepsilon > 0$\uff0c\u4e3a\u4f7f$|x_n-0|<\\varepsilon$\uff0c\u53ea\u9700\u8981<\/p>\n

$$
\n\\frac{|\\cos \\frac{n\\pi}{2}|}{n} < \\varepsilon
\n$$<\/p>\n

$n$\u65e0\u6cd5\u89e3\u51fa\uff0c\u6ce8\u610f\u5230<\/p>\n

$$
\n\\frac{|\\cos \\frac{n\\pi}{2}|}{n} \u2264 \\frac{1}{n}
\n$$<\/p>\n

\u6545\u53ea\u9700\u8981$\\frac{1}{n}<\\varepsilon$\uff0c\u5373$n>\\frac{1}{\\varepsilon}$<\/p>\n

\u8bc1\uff1a<\/p>\n

$\\forall \\varepsilon > 0$\uff0c\u53d6$N=[\\frac{1}{\\varepsilon}]$\uff0c\u5219\u5f53$n>N$\u65f6\uff0c\u5c31\u6709<\/p>\n

$$
\n|x_n-0| = \\frac{|\\cos \\frac{n\\pi}{2}|}{n} \u2264 \\frac{1}{n} \u2264 \\varepsilon
\n$$<\/p>\n

\u4ece\u800c<\/p>\n

$$
\n\\lim \\limits_{x \\to \u221e} \\frac{\\cos \\frac{n\\pi}{2}}{n} = 0
\n$$<\/p>\n

\u4f8b\u4e09<\/h3>\n

\u8bc1\u660e\uff1a<\/p>\n

$$
\n\\lim \\limits_{x \\to \u221e} \\frac{(-1)^n}{(n+1)^2} = 0
\n$$<\/p>\n

\u5206\u6790\uff1a<\/p>\n

$$
\n|x_n-0| = \\frac{1}{(n+1)^2}
\n$$<\/p>\n

$\\forall \\varepsilon > 0$\uff0c\u4e3a\u4f7f$|x_n-0|<\\varepsilon$\uff0c\u53ea\u9700\u8981<\/p>\n

$$
\n\\frac{1}{(n+1)^2} < \\varepsilon
\n$$<\/p>\n

\u5373<\/p>\n

$$
\n(n+1)^2 > \\frac{1}{\\varepsilon}, \\, n > \\frac{1}{\\sqrt{\\varepsilon}}-1
\n$$<\/p>\n

\u8bc1\uff1a<\/p>\n

$\\forall \\varepsilon > 0$\uff08\u4e0d\u59a8\u8bbe$\\forall \\varepsilon < 1$\uff09\uff0c\u53d6$N=[\\frac{1}{\\sqrt{\\varepsilon}}-1]$\uff0c\u5219\u5f53$n>N$\u65f6\uff0c\u6709<\/p>\n

$$
\n|\\frac{(-1)^n}{(n+1)^2}-0| < \\varepsilon
\n$$<\/p>\n

\u5373<\/p>\n

$$
\n\\lim \\limits_{x \\to \u221e} \\frac{(-1)^n}{(n+1)^2} = 0
\n$$<\/p>\n

\u4ee4\u8bc1\uff1a<\/p>\n

$$
\n|x_n-0| = \\frac{1}{(n+1)^2} < \\frac{1}{n+1} < \\frac{1}{n}
\n$$<\/p>\n

\u4e3a\u4f7f$|x_n-0|<\\varepsilon$\uff0c\u53ea\u9700\u8981$\\frac{1}{n}<\\varepsilon$\uff0c\u5373$n>\\frac{1}{\\varepsilon}$\uff0c\u4ece\u800c\u53d6$N=[\\frac{1}{\\varepsilon}]$\u5373\u53ef\u3002<\/p>\n

\u6536\u655b\u6570\u5217\u7684\u6027\u8d28<\/h2>\n

\u5b9a\u7406\u4e00\uff1a\u82e5\u6570\u5217$\\{x_n\\}$\u6536\u655b\uff0c\u5219\u5b83\u7684\u6781\u9650\u552f\u4e00\u3002<\/strong><\/p>\n

\u8bc1\uff1a\u53cd\u8bc1\u6cd5\uff0c\u8bbe$x_n \\rightarrow a, x_n \\rightarrow b$\uff0c\u4e14$a<b$\uff0c\u53d6$\\varepsilon=\\frac{b-a}{2}$<\/p>\n

\u5219\u7531\u6781\u9650\u5b9a\u4e49\u77e5\uff0c\u5bf9\u6b64$\\varepsilon>0$<\/p>\n

$\\exists$\u6b63\u6574\u6570$N_1$\uff0c\u5f53$n>N_1$\u65f6\uff0c$|x_n-a|<\\varepsilon$\uff0c\u5373\u6709$x_n < \\frac{a+b}{2}$
\n$\\exists$\u6b63\u6574\u6570$N_2$\uff0c\u5f53$n>N_2$\u65f6\uff0c$|x_n-b|<\\varepsilon$\uff0c\u5373\u6709$x_n > \\frac{a+b}{2}$<\/p>\n

\u6545\u53d6$N=max{N_1,N_2}$\uff0c\u5f53$n>N$\u65f6\uff0c\u6709<\/p>\n

$x_n < \\frac{a+b}{2}$ \u4e14 $x_n > \\frac{a+b}{2}$\u3002\u77db\u76fe\uff01\u4ece\u800c\u5047\u8bbe\u4e0d\u6210\u7acb\uff01<\/p>\n

\u4e8e\u662f\u7531$a,b$\u7684\u5927\u5c0f\u4efb\u610f\u6027\u53ef\u63a8\u51fa$a=b$\uff0c\u5373\u6781\u9650\u552f\u4e00\u3002<\/p>\n

\u5b9a\u7406\u4e8c\uff1a\u82e5\u6570\u5217$\\{x_n\\}$\u6536\u655b\uff0c\u5219\u6570\u5217$\\{x_n\\}$\u4e00\u5b9a\u6709\u754c\u3002<\/strong><\/p>\n

\u8bc1\uff1a\u8bbe$\\lim \\limits_{x \\to \u221e} x_n = a$\uff0c\u53d6$\\varepsilon=1$\uff0c\u5219$\\exists$\u6b63\u6574\u6570$N$\uff0c\u5f53$n>N$\u65f6\uff0c\u6709$|x_n-a| < 1$\uff0c\u4ece\u800c<\/p>\n

$|x_n| = |(x_n-a)+a| \u2264 |x_n-a| + |a| < 1+|a|$<\/p>\n

\u518d\u53d6 $M=max\\{|x_1|,|x_2|,\u00b7\u00b7\u00b7,|x_N|,1+|a|\\}$<\/p>\n

\u5219\u6709 $|x_n| \u2264 M \\, (n=1,2,\u00b7\u00b7\u00b7)$<\/p>\n

\u6545\u6570\u5217$\\{x_n\\}$\u6709\u754c\u3002<\/p>\n

\u2460 \u82e5\u6570\u5217\u6536\u655b\uff0c\u5219\u6570\u5217\u6709\u754c
\n\u2461 \u82e5\u6570\u5217\u6709\u754c\uff0c\u5219\u6570\u5217\u4e0d\u4e00\u5b9a\u6536\u655b\u3002\u5982\u6570\u5217$\\{(-1)^{n+1}\\}$\u6709\u754c\u4f46\u4e0d\u6536\u655b<\/p>\n

\u5b9a\u7406\u4e09\uff1a\u6536\u655b\u6570\u5217\u7684\u4fdd\u53f7\u6027<\/strong><\/p>\n

\u82e5$\\lim \\limits_{x \\to \u221e} x_n = a$\u4e14$a>0(<0)$\uff0c\u5219$\\exists$\u6b63\u6574\u6570$N$\uff0c\u5f53$n>N$\u65f6\uff0c\u6709$x_n>0(<0)$\u3002<\/p>\n

\u8bc1\uff1a\u53ea\u8bc1$a>0$\u7684\u60c5\u51b5<\/p>\n

\u53d6$\\varepsilon=\\frac{a}{2}$\uff0c\u5219\u7531\u6781\u9650\u5b9a\u4e49\u77e5\uff0c$\\exists$\u6b63\u6574\u6570$N$\uff0c\u5f53$n>N$\u65f6\uff0c\u6709<\/p>\n

$|x_n-a|<\\frac{a}{2}$\uff0c\u5373$\\frac{a}{2}<x_n<\\frac{3a}{2}$<\/p>\n

\u56e0$a>0$\uff0c\u6545$x_n>0$<\/p>\n

\u63a8\u8bba\uff1a\u82e5$\\lim \\limits_{x \\to \u221e} x_n = a$\uff0c\u4e14$\\exists$\u6b63\u6574\u6570$N$\uff0c\u5f53$n>N$\u65f6\uff0c$x_n\u22650(\u22640)$\uff0c\u5219$a\u22650(\u22640)$\u3002<\/p>\n

\u5939\u903c\u51c6\u5219<\/h2>\n

\u82e5$(1) \\, y_n \u2264 x_n \u2264 zn \\, (n=1,2,\u00b7\u00b7\u00b7)$\u53ca$(2) \\, \\lim \\limits<\/em>{x \\to \u221e} yn = \\lim \\limits<\/em>{x \\to \u221e} zn = a$\uff0c\u5219$\\lim \\limits<\/em>{x \\to \u221e} x_n = a$<\/p>\n

\u8bc1\uff1a\u7531\u6761\u4ef6$(2)$<\/p>\n

$\\forall \\varepsilon > 0,\\exists N_1,N_2$<\/p>\n

\u5f53$n>N_1$\u65f6\uff0c$|y_n-a| < \\varepsilon$
\n\u5f53$n>N_2$\u65f6\uff0c$|z_n-a| < \\varepsilon$<\/p>\n

\u4ee4$N=max\\{N_1,N_2\\}$\uff0c\u5219\u5f53$n>N$\u65f6\uff0c\u6709<\/p>\n

$a-\\varepsilon < y_n < a+\\varepsilon, \\, a-\\varepsilon < z_n < a+\\varepsilon$<\/p>\n

\u7531\u6761\u4ef6$(1)$<\/p>\n

$a-\\varepsilon < y_n \u2264 x_n \u2264 z_n < a+\\varepsilon, \\, a-\\varepsilon < z_n < a+\\varepsilon$<\/p>\n

\u5373$|xn – a| < \\varepsilon$\uff0c\u6545$\\lim \\limits<\/em>{x \\to \u221e} x_n = a$<\/p>\n

\u5355\u8c03\u6709\u754c\u6570\u5217\u5fc5\u6709\u6781\u9650<\/h2>\n

\u82e5$x_1 \u2264 x_2 \u2264 \u00b7\u00b7\u00b7 \u2264 xn \u2264 x<\/em>{n+1} \u2264 \u00b7\u00b7\u00b7 \u2264 M$\uff0c\u5219$\\lim \\limits_{x \\to \u221e} x_n = a (\u2264 M)$
\n\u82e5$x_1 \u2265 x_2 \u2265 \u00b7\u00b7\u00b7 \u2265 xn \u2265 x<\/em>{n+1} \u2265 \u00b7\u00b7\u00b7 \u2265 m$\uff0c\u5219$\\lim \\limits_{x \\to \u221e} x_n = b (\u2265 m)$<\/p>\n","protected":false},"excerpt":{"rendered":"

\u6570\u5217\u6781\u9650\u7684\u5b9a\u4e49 \u8bbe$\\{x_n\\}$\u4e3a\u4e00\u6570\u5217\uff0c\u82e5\u6709\u5e38\u6570$a$\uff0c\u5bf9\u4efb\u610f\u7ed9\u5b9a\u7684\u6b63\u6570$\\varepsilon$\uff08\u65e0\u8bba […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[509],"tags":[],"class_list":["post-2129","post","type-post","status-publish","format-standard","hentry","category-mathematics-fundamentals"],"_links":{"self":[{"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/posts\/2129","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/comments?post=2129"}],"version-history":[{"count":0,"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/posts\/2129\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/media?parent=2129"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/categories?post=2129"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.appblog.cn\/index.php\/wp-json\/wp\/v2\/tags?post=2129"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}