6第五章微分中值定理和Taylor展开例5.3.4.设于在（a,+o）中可微，则(1)如果lim·f'()=1,则limf(a)=+00.(2)如果α>0,使得lim (q·f(r) +· f'(r)) =β,则limg (a) = .ar证明.(1）→+o0时，log→+80，故f(a)f'(a)lim2 = lim a·f()=1lim→00 log-n-→00(2)当α>0时,→+o0(→+),故raofra-l.af+a".flim= lim二limf(a)raQ·2Q-1n-001lim(a·(a)+-)==00a$5.4Taylor展开Motivation：我们研究一元函数f的性态(1)如果f(a)在o处连续,则f()-f(ro) =o(1), (→ro)即，在o附近于可用常值函数f(ro）逼近(2)如果f()在o处可微，则f()-[f(ro) + f(ro)(r-o) =o(-o), (→ o),即，在To附近f可用线形函数L逼近L(r) = f(ro) + f'(ro) - (r - ro)(3)如果F"(ro）存在，则由85.2例2f() -[f(ro) + f(ro) (-) +"(ro) (-o)")=o( -o)"), (→ o),即f在ro附近可以用2次多项式逼近.一般地，我们有
6 1ÊÙ ©¥½nÚ Taylor Ðm ~ 5.3.4. f 3 (a, +∞) ¥. K (1) XJ lim n→+∞ x · f 0 (x) = 1, K limn→∞ f(x) = +∞. (2) XJ ∃ α > 0, ¦ limn→∞ (α · f(x) + x · f 0 (x)) = β, K limn→∞ f(x) = β α . y². (1) x → +∞ , log x → +∞, limn→∞ f(x) log x = limn→∞ f 0 (x) 1 x = limn→∞ x · f 0 (x) = 1 (2) α > 0 , x α → +∞ (x → +∞), limn→∞ f(x) = limn→∞ x α · f x α = limn→∞ x α−1 · αf + x α · f 0 α · x α−1 = limn→∞ 1 α (α · f(x) + x · f) = β α . §5.4 Taylor Ðm Motivation: ·ïļê f 5. (1) XJ f(x) 3 x0 ?ëY, K f(x) − f(x0) = o(1), (x → x0), =, 3 x0 NC f ^~¼ê f(x0) %C. (2) XJ f(x) 3 x0 ?, K f(x) − [f(x0) + f 0 (x0)(x − x0)] = o(x − x0), (x → x0), =, 3 x0 NC f ^/¼ê L %C, L(x) = f(x0) + f 0 (x0) · (x − x0) (3) XJ f 00(x0) 3, Kd §5.2 ~ 2, f(x) − [f(x0) + f 0 (x0) · (x − x0) + 1 2 f 00(x0) · (x − x0) 2 ] = o((x − x0) 2 ), (x → x0), = f 3 x0 NC±^ 2 gõª%C. /, ·k
7g5.4Taylor展开定理5.4.1（带Peano余项的Taylor公式).设f在zo处n次可导，则(*) f() =f(ro)+f(ro).(-o)+f"(ro) - (a- zo)? +..()(o) (o)"+o()"), (0)证明.用数学归纳法.n=1.2已经在前面证明.设n=k时（*）成立，则n=k+1时，由假设，fk+1(ro)存在，此时f(zo）在ao处k次可导，由归纳假设-f"(ro) (r - ro)? + ..f'(r) = f'(ro)+f"(ro)-(-ro)+ :21(+1)(ro) (0)+o(o)), (0),!从而由L'Hospital法则f() - [5(0) + f(c0) ( - 20) + .+ ( ( - 20)++)lim(r - ro)*+1) -I(c)+ "(c0) (α-20)++ ( (-0)lim(k + 1)(r- ro)k= 0.即f(a) = f(ro)+f (ro) (-ro)+f"(ro) -(-ro)?+...f(k+1)(ro) - ( - zo)h+1 + o((r - To)*+1), (a → ro).(k + 1)!口从而（*）对n=k+1也成立.由数学归纳法，（*）对任意的n都成立记f(n)(ro)Rn(a) = Rn(ro, z) = f(r) - [f(ro) + f'(ro) (r - ro) + .. +(α-To)"],n!称Rn为Taylor展开的余项.如果f在o处有n阶导数，则Rn(r)=o((r-ro)").(Peano余项)如果条件更强些，则有定理5.4.2(Taylor).设于在开区间(a,b)中有直到n+1阶导数，o,E(a,b)，则日EE(,ro)(或(r,ro))以及（E(ro,)(或(,ro),使得余项可表示为1(n+1)()-(-ro)n+1,(Lagrange余项）Rn(r) = 7(n+1)!以及Ra(1)=f(n+1)(C) (-C)" (μ- zo), (Cauchy 余项)一
§5.4 Taylor Ðm 7 ½n 5.4.1 ( Peano { Taylor úª). f 3 x0 ? n g, K (∗) f(x) = f(x0) + f 0 (x0) · (x − x0) + 1 2!f 00(x0) · (x − x0) 2 + · · · + 1 n! f (n) (x0) · (x − x0) n + o((x − x0) n ), (x → x0) y². ^êÆ8B{. n = 1, 2 ®²3c¡y². n = k (∗) ¤á, K n = k + 1 , db, f k+1(x0) 3, d f 0 (x0) 3 x0 ? k g, d8Bb, f 0 (x) = f 0 (x0) + f 00(x0) · (x − x0) + 1 2!f 000(x0) · (x − x0) 2 + · · · + 1 k! f (k+1)(x0) · (x − x0) k + o((x − x0) k ), (x → x0), l d L’Hospital {K limx→x0 f(x) − [f(x0) + f 0 (x0) · (x − x0) + · · · + f (k+1)(x0) (k+1)! · (x − x0) k+1] (x − x0) k+1 = limx→x0 f 0 (x) − [f 0 (x0) + f 00(x0) · (x − x0) + · · · + f (k+1)(x0) k! · (x − x0) k ] (k + 1)(x − x0) k = 0. = f(x) = f(x0) + f 0 (x0) · (x − x0) + 1 2!f 00(x0) · (x − x0) 2 + · · · + 1 (k + 1)!f (k+1)(x0) · (x − x0) k+1 + o((x − x0) k+1), (x → x0). l (∗) é n = k + 1 ¤á. dêÆ8B{, (∗) é?¿ n Ѥá. P Rn(x) = Rn(x0, x) = f(x) − [f(x0) + f 0 (x0) · (x − x0) + · · · + f (n) (x0) n! · (x − x0) n ], ¡ Rn Taylor Ðm{. XJ f 3 x0 ?k n ê, K Rn(x) = o((x − x0) n ). (Peano {) XJ^r , Kk ½n 5.4.2 (Taylor). f 3m«m (a, b) ¥k n+1 ê, x0, x ∈ (a, b). K ∃ ξ ∈ (x, x0)( ½ (x, x0)) ±9 ζ ∈ (x0, x)( ½ (x, x0)), ¦{L« Rn(x) = 1 (n + 1)!f (n+1)(ξ) · (x − x0) n+1 , (Lagrange {) ±9 Rn(x) = 1 n! f (n+1)(ζ) · (x − ζ) n · (x − x0), (Cauchy {)