02第二講 導(dǎo)數(shù)及其應(yīng)用(34=76)

1、34第二講第二講導(dǎo)數(shù)及其應(yīng)用導(dǎo)數(shù)及其應(yīng)用問題2.1導(dǎo)數(shù)、微分的概念例題1.選擇C，理由如下：函數(shù)在點處可導(dǎo)，即存在()yfx?0x0000()()()limxfxxfxfxx?????????00000000()()()()()()limlimhhfxhfxhfxhfxfxfxhhh???????????????；0000000()()()()limlim()()hhfxhfxfxhfxfxhh????????????????????但

2、是，極限存在函數(shù)在點處可導(dǎo)，000()()limhfxhfxhh???????()yfx?0x例如在點處，，但是函數(shù)在yx?0x?00()()limlim0hhhhfhfhhh????????yx?點處不可導(dǎo).0x?2.選擇B，理由如下：在可導(dǎo)存在??xf0?x0()limxfxx??222000011cos(1cos)1cos(1cos)lim(1cos)limlimlim1cos1coshhhhhfhhfhfhhhhhh??????

3、?????????，故A不對；1cos01()lim2uhufuu?????.10000011(1)1(1)()lim(1)limlimlimlim11hhhhhuehhhhhhhuefeefefufehheheu????????????????????3.【用特例法】選擇D，理由如下：設(shè)10()10.xxfxxx????????，10()10xfxx????????00lim()lim()1xxfxfx????????但是，，在處極限

4、不存在，0lim()1xfx????0lim()1xfx???)(xf0x?故排除A，B，C.4.【涉及不等式條件的極限問題，通常用夾逼準(zhǔn)則】選擇C，理由如下：當(dāng)時，恒有，故)(????x2)(xxf?(0)0(0)0ff???又，20()0(0)lim()0xfxxxfx??????360)(?ag?()()()()()()()limlimxaxagxxgaagxxFaxaxa???????????，[()()]()lim()()xa

5、gxgaxgaaxa????????故是在處可導(dǎo)的充分條件；0)(?ag)(xFax?在處可導(dǎo)存在，)(xFax??()()()()()limxagxxgaaFaxa???????即()()()()()limxagxxgaaFaxa???????()()()()()()()()limxagxxgaxgaxgaaxa??????????存在，()()()()lim[()()]xagxgaxaxgaxaxa???????????0)(?ag

6、故是在處可導(dǎo)的必要條件.0)(?ag)(xFax?5.選擇B，理由如下：，0()0fx??00000()()()limlim0xxxxfxfxfxxxxx????????在可導(dǎo)存在()fx0x00000()()()limlimxxxxfxfxfxxxxx???????（）和（）存在且相等00()limxxfxxx????0?00()limxxfxxx???0?0000()()limlim0xxxxfxfxxxxx?????????.00

7、00000000()()()()limlim0lim0lim0xxxxxxxxfxxxfxfxfxxxxxxxxxxx?????????????????6.12110()cossinxfxxxxx?????????由題設(shè)知存在，故，10001cos()(0)1(0)limlimlimcosxxxxfxfxfxxxx???????????1??，且，故.(0)0f??122000111lim()lim(cossin)limsin0xxxf