部分作业题解答ppt.ppt

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1、第一章第二章作业评讲Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.0证奇偶性应先说明定义域关于原点对称作业讲评Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.P28左极限右极限Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5

2、ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.辨析：()Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.设点x0为f(x)的间断点1）若f(x)在x0的左、右极限都存在,则称x0为f(x)的第一类间断点2）若f(x)在x0的左、右极限至少一个不存在,则称x0为f(x)的第二类间断点+P45,3.判断间断点的类型所以x=-2为第二类间断点P45Evaluationonly

3、.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.P38:EX6辨析：Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.常用近似公式（P71)等价无穷小（P37)x=0处线性化Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.

4、0.0.Copyright2004-2011AsposePtyLtd.P73,6,2)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.方法2：x=1时f（x）=1，点(1,1)在曲线上P73,6,2)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.P73,6,2)解附加题Ev

5、aluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.－2不存在极限存在可导连续７,8,9,10,14Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.－2不存在约去0因子改变了原式的定义域Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5

6、ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.定义2定义3（更常用）Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.()在x0处极限存在在x0处连续Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.下页P53定理2：P5

7、3例7：在x0处可导在x0处连续Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.分段函数的分界点的导数只能按定义求，因为此时的不是同一个表达式NOTE3可导：连续：@EX215EX216附加题Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyr