第一章习题习题1.11.判断下列函数是否相同:①定义域不同;②定义域对应法则相同同;2.解f(0.5)2,f(0.5)10.251.255.解①x21y2,x1y2,0y1yyyycea②lnealn(bxc),eabxc,bxcea,x,ybb6.解①ylnu,uv23,vsinx②yarctanu,u2x35习题1.24.解:①无穷大②无穷小③负无穷大④负无穷大⑤无穷小⑥无穷小5.求极限：⑴lim(3x32x1)3limx32limx12x1x1x12x23lim(x3)1⑵limx2x22x1lim(2x1)5x2atanx⑶lim0xx2x32x212(x1)1limlimlimlimx1x25x4x1(x4)(x1)x1(x4)(x1)x1(x4)(x1)⑷23x25x6(x3)(x2)x31⑸limlimlimx2x24x2(x2)(x2)x2x24x2x2(11x2)⑹limlimx011x2x0(11x2)(11x2)x2(11x2)limlim(11x2)2x0x2x011x21x21⑺limlim3x113xx3x2x2323x32x31⑻limlimx43x52xx2x245313x2x2(x1)(x2)3⑼limlimlim1x1x1x31x1x31x1(x1)(x2x1)3(n1n)(n1n)1⑽lim(n1n)limlim0nn(n1n)nn1n132n1n2⑾limlim1xn2n2n2xn21111112n2⑿lim1lim22222n1nn126.求极限x1⑴limx0tan4x41sin1x⑵limxsinlim1xxx1x1cos2x2sin2x2sinx⑶limlimlim2x0xsinxx0xsinxx0xx⑷lim2nsinxn2narcsinx1y1⑸limlimx02x2y0siny21sin11x2⑹limxsinlimx2sinlim1xx2xx2x1x2k112(k)2(k)⑺lim(1x)xlim(1x)xlim[(1x)x(1x)2]ekx0x0x01x2x1x2⑻limlim1e2xxxx13⑼lim(13tanx)cotx1lim[(13tanx)3tanx(13tanx)1]e3x0x02x33x4422()2⑽lim1lim[(1)23(1)3]e3x3xx3x3x11xx1x⑾lim(13x)xlim(13x)33xlim(13x)3x3xe01xxxx3xx3111⑿limlime1x1xx1x3e11xlim1xx习题1.3(x1)(x1)1、⑴因为函数在x=1点处无定义，f(x)，但是limf(x)2，x=1(x1)(x2)x1点是函数的第一类间断点（可去）。limf(x)，故x=2是二类间断点。x2⑵虽然在x=0处，函数无定义，但是极限存在，故x=0是第一类间断点（可去）⑶limf(x)lim(11)0，limf(x)lim(31)2，极限存在但是不等，x1x1x1x1x=1是第一类间断点。⑷函数在x=0处无定义，上下摆动，极限不存在，是第二类间断点。x2(x3)(x3)(x3)(x21)2、解：f(x)(x3)(x2)(x3)(x2)函数的定义域是：(,3)(3,2)(2,)191841limf(x)，limf(x)，limf(x)