2验证极限limxsi
x存在但不能用洛必达法则得出xx解limxsi
xlim1si
x1极限limxsi
x是存在的xxxxxxxsi
xlim1cosxlim1cosx不存在不能用洛必达法则但limxxxx1
x2si
1x存在但不能用洛必达法则得出3验证极限limx0si
xx2si
1x2si
1x是存在的xlimxxsi
1100极限lim解limx0si
xx0si
xx0si
xx
x2si
12xsi
1cos1xxx不存在不能用洛必达法则但limlimx0si
xx0cosx
1x11xx4讨论函数fxe12e
x0x0
12
在点x0处的连续性
解f0e2limfxlime
x0x0
1
12
e
f0
11l
1x11xx1xlimexx因为limfxlimx0x0x0e
1
l
1xx而lim11l
1x1limx0xxx0x211lim1xlim11x02xx021x2
11l
1x11xx1xlimexx所以limfxlimx0x0x0e1
e
12
f0
因此fx在点x0处连续
8
f习题331按x4的幂展开多项式x45x3x23x4解设fxx45x3x23x4因为
f456f44x315x22x3x421f412x230x2x474f424x30x466f4424
所以
f4f4f44x42x43x442345621x437x4211x43x44fxf4f4x4
2应用麦克劳林公式按x幂展开函数fxx23x13解因为
fx3x23x122x3fx6x23x12x326x23x1230x23x1x23x2fx302x3x23x230x23x12x3302x32x26x3f
4
x602x26x3302x34x6360x23x2
f5x3602x3f6x720f01f09f060f0270f
4
0720f501080f
6
0720
所以
f02f03f404f505f606xxxxx2345619x30x345x330x49x5x6fxf0f0x
3求函数fxx按x4的幂展开的带有拉格朗日型余项的3阶泰勒公式解因为
1f442f4x2
12
1f41x2x444
3x4
132
9
ff43x8
52
x4
3f4x15x216r