高等數(shù)學(xué)1教學(xué)材料33公式

計(jì)算下列極n1

exex2

1bx

xsin 15.x 1ln(1x)tan

8.limarcsinxe x7.x cot(

x x1x ? 10.lim 9.lim(xex)xx0

x0 i 1i211.lima1x2

x…

x a0,i1,2,…,x 1設(shè)f(x)不恒為常數(shù),在[a,b]連續(xù),在(a,b)可導(dǎo),而f(a)=f(b),求證:在(a,b)內(nèi)至少存在一 滿足f()計(jì)算下列極n1n1axm1a xlime

e

2x xsinln(1x) x1

cot(注意分子是 最終仍是2limarcsinxexx0lim(xex)xx0x1x2x?nxxnnx0 1a1a211. na1xaxa1a211. nx ai0,i1,2,…,

lettx412.設(shè)f(x)有二階連續(xù)導(dǎo)數(shù)而 f(0)求證 (x)

f(x)

x

有一階連續(xù)導(dǎo)f

xxf(x)f(x(x)1f(0)x5一、不定式limxxx0limxxxlimcosxsecx

x xx06xxxa

(a 分

1tanx1sin

)x

lim[x11cosx

x2ln(1sin

1x

e變量代 e22271]1 當(dāng) 二.討論函數(shù)f(x)

當(dāng)x在點(diǎn)x0處1lim(1

1x

xex0

x

83.3問解決的方法，使用更高次數(shù)的多項(xiàng)式P(x) a(xx)a(xx)2?a(xx Rn(x)f(x)Pn(9n確定多項(xiàng) 為了讓P(x)在x0附近充分接近f(x)，n 們希望它們?cè)谠擖c(diǎn)yo有有相同的函數(shù)值 yof(點(diǎn)Pnx0fx0有相同的導(dǎo)數(shù)值(相同的切線Pnx0)fx0)；有相同的二階導(dǎo)數(shù)值(相同的彎曲x向Pnx0fx0雖然更高階導(dǎo)數(shù)的含義沒有很直觀的解釋，但是我們還是希望 ((k )f(k)(x)，k0,1,2,3,4, ( n次多項(xiàng)式的更高階導(dǎo)數(shù)全部為0P(x) a(xx)a(xx)2?a(xx P(x)f(x),P(k)(x)f(k)(x k1,2,?,n a0f(x0),1a1f(x0),2!a2f(x0),n!an

((x0

f(k)(k

)(k0,1,2,?,(xP(x)f(x)f(x)(xx)f(x0 (x f(n)(x (x f(k)(x

x0k

k! (x

x0真的很接近嗎？一個(gè)實(shí)f(x)sinx,x0f(2k)(0)0,f(xk1)(0)(1)k1,(1)k

a2k0,a2k1

2k

(x) (x)x ? 7 2n321321---123--yysinyx x-321321--24yx-35 ysinyxx5 6yx ysinyx 4 2--5---(Taylor)中值定理如果函數(shù)fxx0的某個(gè)開區(qū)間(a,b)內(nèi)具有直到(n1)階的導(dǎo)數(shù),則當(dāng)x在(ab)內(nèi)時(shí),fx)可以表示為xx0)的一個(gè)n次多項(xiàng)式與一個(gè)余項(xiàng)Rn(x)之和:f(x)f(

)f(

)(x

)f(x0)(x

x0f(n)(x (x

x0

(f(n1)( 其中Rn(x) (n1)!(xx0 (在x0與x之間Rnx)在(ab)內(nèi)具有直到(n1)階nRn(x0)Rn(x0)Rn(x0)?R(n)(x0)n兩函數(shù)Rnx及xx0n1在以x0x為端點(diǎn) Rn(x) Rn(x)Rn(x0)(xx)n1 (xx)n10 (在x與x之間 (n1)(1x0兩函數(shù)Rx及(n1xx)n在以x及 的區(qū)間上滿足中值定理的條件, nn

(n1)(1

x0)n

(在x與之間0n(n1)(2x 0如此下去,經(jīng)過(n1) 0R(x) R(n1)()(xx)n1 n 0(在x0與n之間,也在x0與x P(n1)(x) R(n1)(x)f(n1)( f(n1)( Rn(x) n

(x

x0 (在x0與x之間0nP(x) f(k)(x0)(xx0nk fx)按xx0的冪展開的n f(k)(xf(x) (xx)kR(nk fx)按xx0的冪展開的nR(x)f(n1)()(x

,x(xx),)n n 日形式的余 f(k)(x f(n1)(f(x)

k

k! (x

x0)kn1!(xx0帶日余項(xiàng)當(dāng)x0 f(k) f(n1)(f(x)k

k n

說n0中值定fx)fx0f(xx0用間接展開得到的求x0的高階導(dǎo)數(shù)值；fx)的n1階導(dǎo)數(shù)恒0fx)為至多n次日余項(xiàng)可以用于近似計(jì)算中估計(jì)誤差f(n1)(x)Mf(n1)(f(n1)(n

(x

x0

n

x

說關(guān)于x0xx0O1帶余項(xiàng)定理如果函數(shù)f(x)在點(diǎn)處有直到n階的導(dǎo)數(shù),f(x)P(x)o((xx)n (xx f(k)(xx0

k

k! (x

x0帶余項(xiàng) 林f(x)f(0)f(0)xf(0)x2 f(n)(0)xn? rn(x)f(x)Pnn只要證 nxx0(xx0 r(n1)(x)f(n1)(x)[f(n1)(x)f(n)(x)(x r(x)r(x)r(x)?r(n1)(x) xx0(xx n(xx n(n1)(xxn000 (x) (x (x) (x? 0xx0n!(xx0 n!(xx0 1f(

(x

)f

(x

n nxx0(xx0術(shù)在0點(diǎn)展開——帶形式余項(xiàng)——局那么，局部林149頁(yè)3.3節(jié)比較簡(jiǎn)單的問題，直接 f(x)arcsinx f(x)f(0)f(0)xf(0)x2f(0)x3o(x3 f(0)arcsin0 f(0) xf(0)1(1x2)3/2(22

x

x(1x2

x

f(0)(1x2)3/2

x(1x2)2(2 x f(x)x1x36直接使用得到的一些常用結(jié)求f(x)ex的n階林解 f(x)f(x)?f(n)(x)ex f(0)f(0)f(0)?f(n)(0)f(n1)(x)e?日形式余項(xiàng)?ex1x

x2

xn

e

(0形式余項(xiàng)

(nex1xx2?xno(xn (x sinxxx3x5?(1)m1

x2m 5!

2msinx(2m1)2 x2m1,(0,1)(2m1)!3sinxx3

?

o(x2n (2n(xcosx1x2x4?

cosx(m1)(2m

x2m2

2mcosx1

?(1)

o(x2n1(x

ln1xx ?(1) n(n1)(1x)n1

,ln(1x)xx

x

?

o(xn(x

(1 1x(1)x2(1)(2) ?(

n1)(1)?(n)(1x)n1xn1,(0,1)(n1)!(1 1x(1)x2(1)(2) ?(

n1)xno(xn(x 11

1xx2

?

(1x)n2

11

1xx2?xno(xn),x間接展開：變量代求函數(shù)的帶有余項(xiàng)2n階林展開fx)e 間接解e

1x

x2

xno(xne

1x2

x4

(x2

o(x2n1x2

x4

(1)nx2n

o(x2n間接展開：先進(jìn)行初等變求sinx4

)的帶余項(xiàng)的n2m林sin(x) 4

(sinxcos x2m sinxx ? (2m

o( cosx1x

x

x

?(1) x2

o(x2m12 4 (2m)!間接展開f(x)x3ln(1x3(xx2x3?(1)n1

o(xn 149頁(yè)3.3節(jié)間接展開：初等變換+代求函數(shù)f(x)1在 1點(diǎn)的三 f(x)1 1(

(x1(x1)(x (x

(1(x利用求極ex22cosx計(jì)算lim x 2解 e2

1x

o(x4cos

1

x2

x4o(x5) 2 e2

2cosx3 2 )

o(x47x4o(x4 原式lim ln(1x) x

x3?

o(xn1x 2

nlim[x

x2ln(1

x

展開至x-2次解ln(1

1)

o[() 2x lim[xx2ln(11 lim[xx2(1 o(1x

2 lim

o(1 x2

1x 習(xí)ln(1xx2)ln(1xx2求lim x xsin(1xx2)(1x)1(1xx2)(1x)1ln(1xx2)ln(1xx2)ln(1x6)ln(1x2o(x5)x2o(x2x2x原式x

o(x2

利 求極 limexsinxx(1 ∵ex1

x2!x3

x x o(x33sinxx

o(x

exsinx

x(1 x1x

o(x3 x

o(x3)x(1 2! 3! 3! lim x

x3

o(x x

x3 利用間接展開的求高階導(dǎo)求f(x)x3sinx的n階局部林展開式并算f(6)(0) (sin

x

sin(0k2 sinxxx3? 3

sin(n3 xn3o(xn3)(n3)!sin(n3f(x)x4

x6

2(n

xno(xnf(6) a6

(0) sinx,x 無(wú)窮次可導(dǎo)已知f(x) 1 x 求f(50解:x0時(shí) 3

x5

x n x2n f(x)

[x

?

o( (2n1

x41x

n

x2n

2n ? (2n

o( 由展開式的唯一性可知f(2n2)(2n

a2n

n

(2n1)!f(2n2)(0)2n2n250nf(50)(0)(1)251 利用確定無(wú)窮小的當(dāng)x0時(shí)確定無(wú)窮小關(guān)于xf(x)ln(1x2)xsin x2

)x(x

o(x4 f(x)cosx 1 x2 1

o(x5)1

2

2!2o(x4) 利用做近似計(jì)算并且估計(jì)精0f(kxk0,1,?n1可以方便計(jì)算出來。方法：f(x)Pn(x)Rn(x)Pn(x)誤差：放大Rn(x)去估計(jì)。0用n=10階計(jì)算e的近似值并估計(jì)誤差ex1x

?

xn

e

(0 (n21x2

?n nee(nRn(x)

xn1(01).(n1)!取x1n10e111

其誤 (n (n3應(yīng)用三階 的近似值并估計(jì)誤3f(x)3x,f(27) f(x)1x3,f(27)1 f(x)

x3,f(27) f

(x)

10x3