高数重要极限证明原创中英文对照版
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标签：&&&&&&&&&&&&&&&高数重要极限证明原创中英文对照版&重要极限Important Limit作者&赵天宇Author:Panda Zhao&&&&&&&我今天想在这里证明高等数学中的一个重要极限：Today I want to prove animportant limit of higher mathematics by myself:650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/wkO7Ul07fPEdYQi2XzguVeGiHFb7dXLdLK4SczU9yZ4!/o/dHu6i.K0NwAA&ek=1&kp=1&pt=0&bo=UQAkAFEAJAADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:width:120height:53" width="120" height="53" border="0" hspace="0" vspace="0" title="" />&想要证明上述极限，我们先要去证明一个数列极限：If we want to give evidence ofthe limit, first of all, there are a limit of a series of numbers according toa certain rule we need to certify:650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/DpZmb7nwhKhYh5ybtCbpsz*Mqz6c7UjW1rUni*1iH84!/o/dE4Y9uEvLQAA&ek=1&kp=1&pt=0&bo=PAAfADwAHwADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-indent:32width:100height:52" width="100" height="52" border="0" hspace="0" vspace="0" title="" />想要证明这个极限，我首先要介绍一个定理和一个法则：Before we begin to prove thelimit, there are one theorem and one rule that are the key point we need to introduce:1.牛顿二项式定理（Binomialtheorem）定理的定义为：Definition of Binomial theorem:650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/64PHY*sasMUm1hySIs89vg1TqLQXEOytuGSx6nfgsx4!/o/dC4sh.K2NwAA&ek=1&kp=1&pt=0&bo=aQAkAGkAJAADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:width:200height:68" width="200" height="68" border="0" hspace="0" vspace="0" title="" />其中&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/nGqWtIpKVk*wYanWuDaXhXxRhpX*lVyM7M5b4ALwCS8!/o/dNkTXOHEEwAA&ek=1&kp=1&pt=0&bo=SQAhAEkAIQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:73height:33" />，称为二项式系数，又有&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/m5mAlWR1QVtf4KDEPTdQJNXfhqdID8CkqRgvWwzt6As!/o/dFSNWuHHEwAA&ek=1&kp=1&pt=0&bo=HgAlAB4AJQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:30height:37" />的记法。Among the formula: we define the&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/nGqWtIpKVk*wYanWuDaXhXxRhpX*lVyM7M5b4ALwCS8!/o/dNkTXOHEEwAA&ek=1&kp=1&pt=0&bo=SQAhAEkAIQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:73height:33" />&as binomialcoefficient, it can be remembered to650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/m5mAlWR1QVtf4KDEPTdQJNXfhqdID8CkqRgvWwzt6As!/o/dFSNWuHHEwAA&ek=1&kp=1&pt=0&bo=HgAlAB4AJQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:30height:37" />.牛顿二项式定理（Binomial theorem）验证和推理过程：The process of the ratiocination of Binomialtheorem:采用数学归纳法We consider to use the&mathematical inductionto solve this problem.当n = 1时(While n = 1:),650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/V0R6ijFwIXUqw9lLlPI0b9*48nUVskaTaqpeRGvXCk0!/o/dOqCXeHEEwAA&ek=1&kp=1&pt=0&bo=9AA2APQANgADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:width:300height:66" width="300" height="66" border="0" hspace="0" vspace="0" title="" />;假设二项展开式在n=m时成立。We can make a hypothesis that the&binomial expansionequation&is true when n = m.设n=m+1，则：So if we suppose that n equal mplus one, we will&&to deduce:650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/LPzRYTxwZqhDkwG6x72Nsy8JA3l*iNJxwerdUseS9CA!/o/dLhvWuHIEwAA&ek=1&kp=1&pt=0&bo=8wBIAfMASAEDACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:300height:405" width="300" height="405" border="0" hspace="0" vspace="0" title="" />&具体步骤解释如下：The specific&step of interpretation :第三行：将a、b乘入；The 3rd&line:&a and b are multiplied into the binomial expansion equation.;第四行：取出k=0的项；The 4th&line: take out of theitem which includes the k = 0 in the binomial expansion equation.;第五行：设j=k-1；The 5th&line: making a hypothesisthat is j = k-1;第六行：取出k=m+1项；The 6th&line: What we need totake out of the item including k=m+1 in the binomial expansion equation.第七行：两项合并；The 7th&line: Combining the twobinomial expansion equation.第八行：套用帕斯卡法则；The 8th&line: At this line weneed to use the Pascal’s Rule to combine the binomial expansion equation whichare650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/6zcEU2pulWnDvaBLU6GomaRL9kwpjRTU8C1r3Ptbf7U!/o/dGZsVOHIEwAA&ek=1&kp=1&pt=0&bo=ugAnALoAJwADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:186height:39" />.;&接下来介绍一下帕斯卡法则(Pascal’s Rule)。So at this moment, we should get someknowledge about what the Pascal’s Rule is. Let’s see something about it:帕斯卡法则(Pascal’s Rule)：组合数学中的二项式系数恒等式，对于正整数n、k(k&=n)有：Pascal’s Rule: a binomial coefficientidentical equation of combinatorial mathematics. For the positive integer n andk (k&=n), there is a conclusion:&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/GzLglhLSSKb0XXcyM5ouDhRu9czDA.haCgqkzidkyb4!/o/dDKri.K4NwAA&ek=1&kp=1&pt=0&bo=pAAnAKQAJwADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:width:340height:81" width="340" height="81" border="0" hspace="0" vspace="0" title="" />& & & & & & & & &&通常也可以写成：& & & & & & & & &&There is also commonly written:650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/Yu4sSMsIQOggdns2LEI0D4oqn5pU3alcF.BClPS4cwY!/o/dMsBXOHKEwAA&ek=1&kp=1&pt=0&bo=igAnAIoAJwADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:width:300height:85" width="300" height="85" border="0" hspace="0" vspace="0" title="" />&代数证明：Algebraic proof:重写左边：We can rewrite the left combinatorial item:650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/rOiCQodKNvVwqihvkzyG9N69rB7i*QFM2J0D8vCP2pk!/o/dK22iOKzNwAA&ek=1&kp=1&pt=0&bo=yQBDAMkAQwADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:300height:100" width="300" height="100" border="0" hspace="0" vspace="0" title="" />通分；reductionof fractions to a common.650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/qCsfvnyzH8qnyZTGCHwnKIPvcuZrqbgEPOntCq51AXc!/o/dK9*8eEtLQAA&ek=1&kp=1&pt=0&bo=SAAhAEgAIQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:100height:46" width="100" height="46" border="0" hspace="0" vspace="0" title="" />&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&合并多项式；combining the polynomial.650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/VttZYsFVbTucLbOh6UpmBQm3.YqawKz7dIAsYX6o3a8!/o/dFJ3guK2NwAA&ek=1&kp=1&pt=0&bo=hQAnAIUAJwADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:200height:59" width="200" height="59" border="0" hspace="0" vspace="0" title="" />& & & & & & & & & & & & &&证明完成；The Pascal’s Rule has been proved.接下来只要要证明650) this.width=650;" alt="图片" src="/psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/GbVl*kMAoBjY8Ggi.sISxbZZ9BNSMsjmHaPsdsCCu44!/b/dE4Y9uEvLQAA&ek=1&kp=1&pt=0&bo=PAAfAAAAAAADAAY!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:60height:31" />是单调增加并且有界的，那么就可以得到它存在极限，我们通常称它的极限为e。So what is our next step? The progression ofnumbers according to a certain rule of&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/H2ANXIs1CUIZT2RU8yXqhyu9Tb15cp4J99VpcHHvQ0Y!/o/dJnwUuHHEwAA&ek=1&kp=1&pt=0&bo=DgARAA4AEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:14height:17" />should be proved that it is a monotonicincrease sequence and has a limitation. If we can do these things, we will drawa conclusion that the sequence has an limitation which we generally call e.650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/LuqK0WaRzYn3ubVFdVLd9JEbakaBBCEx0cCweXqYEaY!/o/dCYQ7eEyLQAA&ek=1&kp=1&pt=0&bo=QAFkAEABZAADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:width:400height:125" width="400" height="125" border="0" hspace="0" vspace="0" title="" />&类似的，我们可以得到：We can analogously get the650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/zJLY3aFlLNu.wYH2YbYzIxt*8NPPjH5dU*C11CxJbnA!/o/dJuHWuHKEwAA&ek=1&kp=1&pt=0&bo=FAARABQAEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:20height:17" />:&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/MasNaBuMATjuNEf7TD.0zviZqRbC80dXeg*E1ZU4XVs!/o/dKiDguK2NwAA&ek=1&kp=1&pt=0&bo=lQFmAJUBZgADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:width:500height:126" width="500" height="126" border="0" hspace="0" vspace="0" title="" />可见，&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/H2ANXIs1CUIZT2RU8yXqhyu9Tb15cp4J99VpcHHvQ0Y!/o/dJnwUuHHEwAA&ek=1&kp=1&pt=0&bo=DgARAA4AEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:14height:17" />&和650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/zJLY3aFlLNu.wYH2YbYzIxt*8NPPjH5dU*C11CxJbnA!/o/dJuHWuHKEwAA&ek=1&kp=1&pt=0&bo=FAARABQAEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:20height:17" />相比，除了前两个1相等之外，后面的项都要小，并且650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/zJLY3aFlLNu.wYH2YbYzIxt*8NPPjH5dU*C11CxJbnA!/o/dJuHWuHKEwAA&ek=1&kp=1&pt=0&bo=FAARABQAEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:20height:17" />多一个值大于0的项目，因此：Thus it can be seen, comparing&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/H2ANXIs1CUIZT2RU8yXqhyu9Tb15cp4J99VpcHHvQ0Y!/o/dJnwUuHHEwAA&ek=1&kp=1&pt=0&bo=DgARAA4AEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:14height:17" />&&with&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/zJLY3aFlLNu.wYH2YbYzIxt*8NPPjH5dU*C11CxJbnA!/o/dJuHWuHKEwAA&ek=1&kp=1&pt=0&bo=FAARABQAEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:20height:17" />&, all of the items of the&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/H2ANXIs1CUIZT2RU8yXqhyu9Tb15cp4J99VpcHHvQ0Y!/o/dJnwUuHHEwAA&ek=1&kp=1&pt=0&bo=DgARAA4AEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:14height:17" />&&are lower thanthese items in&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/zJLY3aFlLNu.wYH2YbYzIxt*8NPPjH5dU*C11CxJbnA!/o/dJuHWuHKEwAA&ek=1&kp=1&pt=0&bo=FAARABQAEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:20height:17" />&except the 1stand the 2rd one are equaling. In addition it has an item whose value is biggerthan zero that is in the&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/zJLY3aFlLNu.wYH2YbYzIxt*8NPPjH5dU*C11CxJbnA!/o/dJuHWuHKEwAA&ek=1&kp=1&pt=0&bo=FAARABQAEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:20height:17" />. So we can get a point :650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/00NT0Po3cCo1G0F0wWRtSQzcQZSvagNaR8jLSQlEsIA!/o/dK0giuKvNwAA&ek=1&kp=1&pt=0&bo=KgARACoAEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:width:42height:17" />所以数列是单调递增的得证，接下来证明其有界性：Because of the point, we can prove thesequence is an monotonic increase sequence, so we remain only one thing shouldbe proved that is the sequence’s limitation. So let’s get it :650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/CKE24yb1H0aBSNjPA.hgli6lV72YruDJjC5mS.byQA0!/o/dDgmh.KsNwAA&ek=1&kp=1&pt=0&bo=JAFaACQBWgADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:width:400height:123" width="400" height="123" border="0" hspace="0" vspace="0" title="" />&可见{&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/H2ANXIs1CUIZT2RU8yXqhyu9Tb15cp4J99VpcHHvQ0Y!/o/dJnwUuHHEwAA&ek=1&kp=1&pt=0&bo=DgARAA4AEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:14height:17" />&}是有界的，所以根据数列极限存在准则可得：Thus it can be seen , the sequence of&650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/H2ANXIs1CUIZT2RU8yXqhyu9Tb15cp4J99VpcHHvQ0Y!/o/dJnwUuHHEwAA&ek=1&kp=1&pt=0&bo=DgARAA4AEQADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:text-align:width:14height:17" />&has a limitation , as we know, we can draw aconclusion by the means of the rule of limitation of sequence exiting:650) this.width=650;" alt="图片" src="http://r.photo./psb?/a4e06645-d9aa-49cf-9a0c-a8de035fa6b5/1qresYrV.DozGHvbh2ugdQdfJSDNaU871jiDObSVerQ!/o/dB8Kh.K4NwAA&ek=1&kp=1&pt=0&bo=SAAfAEgAHwADACU!&su=&sce=0-12-12&rf=2-9" style="margin:0padding:0border-width:0border-style:vertical-align:width:144height:62" width="144" height="62" border="0" hspace="0" vspace="0" title="" />本文出自 “” 博客，请务必保留此出处标签：&&&&&&&&&&&&&&&
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[考研资料]高数第五版1-6极限存在准则+两个重要极限.ppt 37页
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函数与极限 三、小结 下页 下页 1.夹逼准则 证 一、极限存在准则 准则Ⅰ
如果数列xn,yn及zn满足下列条件: 那末数列xn的极限存在, 且. 取N=max{N1, N2},上两式同时成立, 上述数列极限存在的准则可以推广到函数的极限 n N1时 n N2时 注意: 准则Ⅰ和准则Ⅰ'称为夹逼准则. 准则Ⅰ′ 如果当x∈U0(x0,δ)(|x| M)时有 那末 例1 解 由夹逼定理得 例2 求 夹逼定理 解 2.单调有界准则 单调递增 单调递减 单调数列 几何解释: 准则Ⅱ
单调有界数列必有极限. 例3
极限存在 证 (舍去) 显然{xn}单调递增,只需再证{xn}有界,用数学归纳法. 假设xk 3 {xn}有界,极限存在. 数列{xn}极限存在的充要条件是: 对?ε 0,?正整数 N ,s.t. m N,n N时 证明(略) . 3*. 柯西极限存在准则(柯西审敛原理)
二、两个重要极限 首先看看在计算机上 进行的数值计算结果： 0. 0.1 0. 0.01 0. 0.001 0. 0.... 0. 0. 证 则 且 一般地 其中, a ≠0 为常数. 求 例4 解 求 例5 解 求 例6 解 求 例7 解 练习题
求 (1) 解 (2) 定义 定义 显然xn+1 xn,{xn}单调递增 有界,极限存在,记为 一般地 其中,
k≠ 0 为常数. 求 例8 解 求 例9 解 求 例10 解 求 例11 解 ( 1? ) 常用的方法 求 例12 解 ( 1? ) 求 例13 解 ( 1? ) 1.两个准则 2.两个重要极限 夹逼准则;
单调有界准则 . * *
5、利用极限存在准则证明数列的极限存在，并求出该极限 .一、1、；
二、1、2；
正在加载中，请稍后...2016考研高数：极限中的“极限”(二)
前面我们已经介绍了等价无穷小替换公式的应用及注意事项，接下来，跨考教育数学教研室佟老师为大家继续说说极限的计算方法。
极限的第三种方法就是洛必达法则。首先，要想在极限中使用洛必达法则就必须要满足洛必达法则，说到这里有很多同学会打个问号，什么法则，不就是上下同时求导?其实不尽然。
洛必达有两种，无穷比无穷，零比零，分趋近一点和趋近于无穷两种情况，以趋近于一点来说明法则条件，
条件一：零比零或者无穷比无穷(0/0，∞/∞);条件二：趋近于这一点的去心领域内可导，且分母导数不为零;条件三：分子导数比分母导数的极限存在或者为无穷，则原极限等于导数比的极限。
在这里要注意极限计算中使用洛必达法则必须同时满足这三个条件，缺一不可，特别要注意条件三，导数比的极限一定是存在或者为无穷，不能把无穷认为是极限不存在，因为极限不存在还包括极限不存在也不为无穷这种情况，比如：x趋近于零，sin(1/x)的极限不存在也不为无穷。每次使用都必须验证三条件是否同时满足。
再来看看重要极限，重要极限有两个，一个是x趋近于零时，sinx/x趋近于零，另一个是x趋近于零时，(1+x)1/x趋近于e，或者写成x趋近于无穷，(1+1/x)x趋近于e(1∞形式)，总结起来就是(1+无穷小量)无穷小量的倒数，所以要记住重要极限的特点，并可以将其推广，即把x换成f(x)，在f(x)趋近零，sinf(x)/f(x)趋近于零，(1+f(x))1/f(x)趋近于e，或f(x)趋近无穷，(1+1/f(x))f(x)趋近于e，还要注意当给你幂指函数的极限计算，先要判断他是不是1∞形式，如果是，就可以考虑利用重要极限解决，凑出相应的形式就可以得出结论。
这里还要特别的提一下几个未定式(∞-∞，0·∞，1∞，00，∞∞)，这五个未定式需要转化为0/0或∞/∞,其中∞-∞可以通过通分、提取或者代换将其转化，0·∞可以将0或者∞放在分母上，以实现转化，1∞，00，∞∞利用对数恒等变化来实现转化，其中1∞还可以利用重要极限计算。
综上所述，等价无穷小替换和重要极限要掌握基本公式和推广，可以将任意变形公式转化为标准形式，并且给定一个极限首要任务就是利用等价无穷替换公式化简。洛必达法则处理七种未定式，灵活地将不同形式的极限转化为0/0或∞/∞，计算时注意满足洛必达法则的三个条件，希望同学们可以掌握基础，灵活地解决不同类型的极限。
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