微积分英文课件 .ppt

1、00lim()()xxf xf xExample224422322 231lim(2)323 22xxxfxx 一、is continuous at0 x()f xLimitsa.0000,0,1,0,0.0二b.00311lim1xxxExample22limxaxaxaxa301tan1 sinlimxxxxc.1sinlim0 xxx0tan6limsin2xxx00Examplelim2 sin(0)2nnnxx221sinlim21xxxxsinlim0 xxx01lim sin0 xxxd.11021lim21xxxExamplelim1xxxxx3sinlim32cosxxxxx

2、e.Example2lim()xxxx111lim()ln1xxxlim(ln)xxxlnlimxxxxExamplelim()xxxxxf.0Example12sinlim2xxxxg.exxx )11(lim13101tanlim()1 sinxxxxExample3lim()6xxxx1limtan()4nnn232limcos11xxxxxj.Examplek.Example40162lim(16)hhfhsinlim0 xxx20limsin0 xxx4)(xxf)sin1(sinlim)1(xxxxxsin1sin)1(21cos21sin2xxxx21cos)1(21sin2xx

3、xx无穷小有界机动 目录 上页 下页 返回 结束 Exampleexists?If so,find the value of and the value of the limit.133lim222xxaaxxxaIs there a number such thatSolutionBecausea0)1(lim22xxxaaaxxx15)33(lim22ExampleIf we know that the limit015a133lim222xxaaxxxdoes not exist,it follows that the value of must be 0,namely,a1515aso

4、 1)2)(1()2)(3(3lim1315153lim133lim2222222xxxxxxxxxxaaxxxxxTrue or false1.4442828lim()limlim4444xxxxxxxxx2.2212211lim(67)67lim56lim(56)xxxxxxxxxxx3.12211lim(3)3lim24lim(24)xxxxxxxxx4.If5.5lim()0 xf x5lim()0,xg xthen does not exist。5()lim()xf xg xIf5lim()xf x 5lim(),xg x then does not exist。5()lim()xf

5、 xg xTrue or falseandand6.If exists,7.2lim()()xf x g xThen the limit must be 。If0lim()xf x 0lim(),xg x then0lim()()0.xf xg x(2)(2)fglimxxxExampleTrue or falseand8.If exists9.2lim()xafxthen exists。If(1)0(3)0,fand fthen there exists a number between 1 and 3 such that clim()xaf x()0.f c True or false10

6、.If()1f xfor all xand0lim()xf xthen0lim()1.xf xTrue or falseexists,三、Choose the best answer for each of the Following,322sin3lim0 xkxx1.Ifthenk=_.,1)1(lim10ekxxx2.Ifthenk=_.94.32.23.1.DCBA2.2.1.1.DCBA四、四、Fill in the blanks:sin1.lim()xxxxe0112.lim(sinsin2)2xxxxx03.lim12xxx104.limxxeExample If)(xf,2)c

7、os1(xxa0 x,10 x,)(ln2xb0 xis continuous at x=0,then a=,b=.20)cos1(lim)0(xxafx2a)(lnlim)0(20 xbfxblnbaln122e机动 目录 上页 下页 返回 结束)1)()(xaxbexfx,0 xFind a and b.机动 目录 上页 下页 返回 结束 Example Ifhas an infinite discontinuity at,1xand has a removable discontinuity at0 xSolution)1)(lim0 xaxbexxbexaxxx)1)(lim0ba10

8、1,0ba,1x)1(lim1xxbexxSo,does not exist.0)(lim1bexxeebxx1lim机动 目录 上页 下页 返回 结束 Because has an infinite discontinuity at ,soBecause has a removable discontinuity atffExample Find a,b,such that0)1(lim33bxaxx0)1(lim313xbxxa0)1(lim33bxaxx机动 目录 上页 下页 返回 结束 SolutionSincethus01lim33xbxaxx331xy,01a,1a)1(lim33

9、xxbx2333231)1(1limxxxxx0 xy机动 目录 上页 下页 返回 结束 therefore)1)(1(sin)1()(xxxxxxf机动 目录 上页 下页 返回 结束 Example Find the points of discontinuity and identify its type,where)1)(1(sin)1(lim1xxxxxx1sin21 x=1.)(lim1xfx x=1.,1)(lim0 xfx,1)(lim0 xfx x=0.机动 目录 上页 下页 返回 结束 has a removable discontinuity atfffhas a jump

10、 discontinuity athas an infinite discontinuity at Example Find.sin12lim410 xxeexxxSolutionxxeexxxsin12lim410 xxeeexxxxsin12lim43401xxeexxxsin12lim410 xxeexxxsin12lim4101机动 目录 上页 下页 返回 结束.1sin12lim410 xxeexxx机动 目录 上页 下页 返回 结束.)321(lim1xxxxLetxxxxf1)321()(xxx11)()(33231Then)(xf3x133.3)(limxfxExample F

11、indUsing the Squeeze Theorem we get六、六、ExampleShow that there is a root of the equationxex2between 0 and 1.SolutionLetWe have01)0(g01)1(1egThus N=0 is a number between g(0)and g(2).Since g is continuous on 0,1,so the IntermediateValue Theorem says there is a number c between 0 and 1 such that g(c)=0

12、.That is ,)(2xexgx.2cec六、六、ExampleProve that the equation has exactly one real root。xx sin1.If,then=_.hxfhxfh)()3(lim0001)(0 xfFill in the blanks:2.If,then=_.hxfhxfh)()(lim0001)(0 xf3.If,then=_.hhxfhxfh)3()2(lim0001)(0 xfFill in the blanks:4.If,then=_.hhxfhxfh)()(lim0001)(0 xfxxfxfx)3()2(lim05.If,th

13、en=_.hbhxfahxfh)()(lim0006.Ifthen,1)0(0)0(fandf1)(0 xf=_.Fill in the blanks:5.If is continuous at and,then=_.)1(f,1x)(xfFill in the blanks:21)(lim1xxfx1)0(1)0(fandfxxfexx1)(lim07.If,then=_.1)1(3)3(lim.81xaxIfxx,thena=_.Derivatives一、一、The chain rule Example Differentiate the function)(sin.3arctan)(.2

14、)1ln()(.12nnnxfyxxxfxxxxf0001sin)(.5.42xxxxxfxxyxaaxExampleFind.)(,)(),(xffxffxff,sin)21(xxf1.if 二、二、Implicit DifferentiationExample1.Find if 22dyxd1xyeyx2.Find and if)0(y 823yxydydxExample1.Find if)(nfxxxfsin)(三、三、Higher Derivatives2.Find if)(nf(0).cxdyadbcaxbExample1.Find if f xyyx四、四、logarithmic

15、differentiation2.Find if yxxysin五、五、Differential()dyfx dxFind if dyxaaxxxyExampleLet 0,0,1sin)(2xbaxxxxxfFind the values of a and b that make differentiable everywhere.()f xBecause f is differentiable everywhere,thus f is continuous and differentiable at x=0ExampleSolutionLet 0,0,1sin)(2xbaxxxxxfFin

16、d the values of a and b that make differentiable at .()f xBecause f is differentiable at ,thus f is continuous at x=0.ExampleSolution0 x 0 x Since f is continuous at x=0200001lim()limsin0lim()lim()(0)xxxxf xxxf xaxbbfbThus0b Since f is differentiable at0 x 2000001sin()(0)1(0)limlimlimsin000()(0)(0)limlim00 xxxxxxbf xfxfxxxxf xfaxbbfaxx0a Thus