2022年極限知識(shí)點(diǎn)

1、高中數(shù)學(xué)第十三章魂 高中數(shù)學(xué)第十三章魂限考試內(nèi)容： 考試內(nèi)容：教學(xué)歸納法.數(shù)學(xué)歸納法應(yīng)用.數(shù)列日勺極限.函數(shù)日勺極限.根限日勺四則運(yùn)算.函數(shù)日勺持續(xù)性.考試規(guī)定： 考試規(guī)定：(1) 理解數(shù)學(xué)歸納法勺原理，能用數(shù)學(xué)歸納法證明某些簡(jiǎn)樸勺數(shù)學(xué)命題.(2) 理解數(shù)列極限和函數(shù)極限勺概念.(3) 掌握極限勺四則運(yùn)算法則；會(huì)求某些數(shù)列與函數(shù)勺極限.(4) 理解函數(shù)持續(xù)勺意義，理解閉區(qū)間上持續(xù)函數(shù)有最大值和最小值勺性質(zhì).§13.極限知識(shí)要點(diǎn)

2、 極限知識(shí)要點(diǎn)1. ⑴第一數(shù)學(xué)歸納法：①證明當(dāng)n取第一種n0時(shí)結(jié)論對(duì)勺 ②假設(shè)當(dāng)n = k ( k G N + ,k > n0 ) 時(shí)，結(jié)論對(duì)日勺，證明當(dāng)n = k +1時(shí)，結(jié)論成立.⑵第二數(shù)學(xué)歸納法：設(shè)P(n)是一種與正整數(shù)n有關(guān)日勺命題，如果① 當(dāng) n = n0 ( n0 G N + )時(shí)，P(n)成立；② 假設(shè)當(dāng)n n°)時(shí)，P(n)成立，推得n = k +1時(shí)，P(n)也成立.那么，根據(jù)①②對(duì)一切自然數(shù)n &g

3、t; n0時(shí)，P(n)都成立.2. ⑴數(shù)列極限日勺表達(dá)措施：① lim a = anT8② 當(dāng)n — 8時(shí)，a — a .⑵幾種常用極限：① lim C = C ( C為常數(shù))n—8② lim — = 0 (k G N, k是常數(shù))n—8 nk③ 對(duì)于任意實(shí)常數(shù)，特別地，如果C是常數(shù)，那么lim (C - f (x)) = C lim f (x).xT x° x T x°lim [ f (x)]n = [ lim f

4、 (x)]n ( n e N + )xT x x T x0 0注：①各個(gè)函數(shù)日勺極限都應(yīng)存在.②四則運(yùn)算法則可推廣到任意有限個(gè)極限日勺狀況，但不能推廣到無(wú)限個(gè)狀況⑶幾種常用極限：① lim上=0nTs x② lim ax = 0 (01)xT+8 xT—s③ lim 耍=1 n lim — = 1xT0 x xT0 sin x1 1④ lim(1 + —)x = e , lim(1 + x)x = e ( e = 2.71828183

5、)x Ts x xT04. 函數(shù)勺持續(xù)性：⑴如果函數(shù)f(x), g(x)在某一點(diǎn)x = x持續(xù)，那么函數(shù)f (x) ± g (x), f (x) - g (x),f^ (g (x)尹0) 0g (x)在點(diǎn)x = x0處都持續(xù).⑵函數(shù)f (x)在點(diǎn)x = x0處持續(xù)必須滿足三個(gè)條件：①函數(shù)f (x)在點(diǎn)x = x0處有定義；②lim f (x)存在；③函數(shù)f (x)在點(diǎn)x = x0處勺極限值xTx0等于該點(diǎn)日勺函數(shù)值，即lim

6、 f (x) = f (x0).xTx0⑶函數(shù)f (x)在點(diǎn)x = x0處不持續(xù)(間斷)日勺鑒定：如果函數(shù)f (x)在點(diǎn)x = x0處有下列三種狀況之一時(shí)，則稱x0為函數(shù)f (x)日勺不持續(xù)點(diǎn).①f (x)在點(diǎn)x = x0處沒(méi)有定義，即f (x0)不存在；②lim f (x)不存在；③lim f (x)存在，x T x 0 x T x 0但 lim f (x)尹 f (x0).xTx05. 零點(diǎn)定理，介值定理，夾逼定理：⑴零點(diǎn)定理：設(shè)函