经济高数课件1-4

第四节极限运算法则三、求极限方法举例一、极限运算法则二、复合函数的极限运算法则定理1一、极限四则运算法则,)(lim,)(limBxgAxf设.0,)()(lim)3(BBAxgxf其中;)]()(lim[)2(BAxgxf;)]()(lim[)1(BAxgxf则).(lim)](lim[,,)(lim1xfcxcfcxf则为常数而存在如果推论常数因子可以提到极限记号外面..)]([lim)](lim[,,)(lim2nnxfxfnxf则是正整数而存在如果推论定理2,lim,lim}.{}{ByAxyxnnnnnn和有数列设.lim0),2,1(0)3(BAyxBnynnnn时，且当;)(lim)1(BAyxnnn那末;lim)2(BAyxnnn二、复合函数的极限运算法则定理3.)(lim)]([lim,)(lim,)(),(,)(lim00100AufxfAufaxxUxaxauxxauoxx则又内的某去心邻域但在若.531lim1232xxxx求例解)53(lim22xxx5lim3limlim2222xxxxx5limlim3)lim(2222xxxxx52322,3531lim232xxxx)53(lim1limlim22232xxxxxx.373123三、求极限方法举例小结则有设,)(.1110nnnaxaxaxf)lim()lim()(lim110000nnxxnxxxxaxaxaxfnnnaxaxa10100).(0xf则有且设,0)(,)()()(.20xQxQxPxf)(lim)(lim)(lim000xQxPxfxxxxxx)()(00xQxP).(0xf.,0)(0则商的法则不能应用若xQ解22112lim.23xxxx例求.,,1分母的极限都是零分子时xx-1.先约去不为零的因子后再求极限)1)(3()1)(1(lim321lim1221xxxxxxxxx31lim1xxx.21解132lim.1xxx例3求.,,1分母的极限都是零分子时x1.x先约去不为零的因子后再求极限1132(32)(32)limlim1(32)(1)xxxxxxxx11lim32xx1.4011lim1.12xxx解011lim.xxx练习求0011(11)(11)limlim(11)xxxxxxxx01lim11xx1.2011lim1.1nxxxn一般地.147532lim42323xxxxx求例解.,,分母的极限都是无穷大分子时x.,,3再求极限分出无穷小去除分子分母先用x147532lim2323xxxxx.7233147532limxxxxx.147532lim5232xxxxx求例.,,分母的极限都是无穷大分子时x.,,3再求极限分出无穷小去除分子分母先用x147532lim232xxxxx.0332147532limxxxxxx.147532lim62324xxxxx求例.,,分母的极限都是无穷大分子时x.,,4再求极限分出无穷小去除分子分母先用x147532lim2324xxxxx.4242147532limxxxxxx解小结为非负整数时有和当nmba,0,000nnnmmmxbxbxbaxaxa110110lim,,,,0,,00mnmnmnba当当当).21(lim7222nnnnn求例解．是无穷小之和时,n)21(lim222nnnnn2)1(21limnnnn)11(21limnn.21先变形再求极限.221limnnn28lim(34).xxxx例求解2222(34)(34)lim(34)lim(34)xxxxxxxxxxxxxx223434/limlim(34)(13/4/1)xxxxxxxxx2lim(34)3/2.xxxx).(lim,0,10,1)(902xfxxxxxfx求设例解,,0两个单侧极限为是函数的分段点x)(lim0xfx,1)(lim0xfx,1左右极限存在且相等,.1)(lim0xfx故)1(lim0xx)1(lim20xxyox1xy112xy