內(nèi)容案例教案

第十六次1 (1)lim x100exx1

x tlim x100

x2 lim

x

t

t100 a(2)lim

bxcx

a0,b0,c

1x0 1 axbxcx33x 原式lim3x a bxcx31=explim x a

xb

c 1=explim

lim lim x 3

x 3 x 3 xln xln xlnc=explim

lim lim x 3

x 3

x

3 =exp

3(lnalnblnc)

=abc3

Previous 1(3)lim axbxcxx1

x a0,b0,c11amaxabc111axx1

axbxc

x

3axx1axx 1lim

3axx

alim 3 1x x111 lim axbxcx x

max{a,b,cPrevious

lim

esin sinxexcosxxcos3x (ett (sinxxcosx解原式lim txlim

x(1)e(sinxxcosxx xlim(1)e

limsinxxcos x3

xPrevious 第十七次求函數(shù)f(x) xe(帶Lagrange余項(xiàng)

n階Maclaurin公解fnx)xexnC0(ex)(n)xC1(ex)(n1)(x ex(nxf(n)(0)f(x) f(0) f(0)x 2

f(0)x2...

(n)(0)x

f(n1)()xnn (n1)

Previous f(x) f(0) f(0)x 2

f(0)x2...

(n)(0)x

f(n1)()xnn (n1)xx2 x ... x2 (n1)(n1)e(n1)

xnPrevious 求函

f(x) x

x03帶Lagrange余項(xiàng)的四 公式1(n n解f(n)(x) (1)x

xnf(x) f(3) f(3)(x3)

f(3)(x3)2 f(3)(x3)33 f(5)()(x3)5

f(4)(3)(x3)413

(x3) (x3)2 3

(x3) (x3)4

(x3)

Previous 第十八次 fx)

x

在基x01帶Peano余項(xiàng)的n 公式fx)

3x

311(x13 x1 1

o(x1)n3k

PreviousNext5

x 1x1lim1x1x x2(cosxe 1x 11x211x

1x81

o(x4 1 x o(x4 cosxe 1 x2o(x2)(1x23x2o(x2)

o(x2))PreviousNext1x1x1x1lim x x2(cosxe 1x

o(x4lim x

x22

x2o(x2) 1x

o(x4lim x

3x4o(x4 122PreviousNext設(shè)f(x)在x=a點(diǎn)連續(xù)，f(a)0 f(a)01lim

f(x

x x

f(a)(xa)解法一f(x) f(a) f(a)(xa) 2

f()(xa)1f(a)(xa)

f()(xa)

x原式lim 2 x

f(a)(xa 1f()(xa)xlim1 x

2f(a

PreviousNext1 f()(xa)xlim1 x

2f(a f()(xa) explim

2f(a xax

f()explim

2f(a)exp

f(a) 2f(a) PreviousNext解法二：lim

f(x x x

fa)(xa f(x explim 1 x

f(a)(xa xaf(x) f(a)(xa)explim x