第五章微分中值定理和Taylor展开85.1函数极值85.2中值定理定理5.2.1(Rolle).设函数f在[a,b]上连续，在(a,b)内可微，且f(a)=f(b),则存在≤E(a,b),使得f"(E)=0.证明.连续函数f在闭区间[a,b]上可以取到最大值M和最小值m.如果M=m，则f恒为常数，从而f'=0;如果M>m，则由f(a)=f(6)知m与M中至少有一个是被f在内点εE(a,b)处所取得，由Fermat定理，f(s)=0.口定理5.2.2(Lagrange)。设函数于在[a,b]上连续，在（a,b)内可微，则存在EE(a,b),使得F() = =1(), 或 () (a) = f()( - a),b-a证明.令F(n)= (m) -[(a)+ 二(a-a) b-n则F(a)=F(b)=0, F满足定理1的条件.从而日EE(a,b),使得 F(s)=0. 此口即为满足定理要求的6.定理5.2.3(Cauchy).设函数f,g在[a,b]上连续，在（a,b)内可微，且g(r)≠0,VE(a,b).则存在E(a,b),使得f(b) - f(a) -f'(E)g(b) -g(a)g ()证明.由定理1和g ≠0知 g(b)≠g(a).令F(1) = f(a) -[9(a)+ ()-((g(g() - g(a))g(b) - g(a)则F(a)=F(b)=0,F满足定理1的条件,从而日EE(a,b),使得F()=0.E即口为满足要求的点注：令g(a)=，则Cauchy定理可以推出Lagrange定理。以上三个定理均称为微分中值定理定理5.2.4(积分第一中值定理).设f,g在[a,b]上Riemann可积，且g(a)不变号，则日，inf,f()≤μ≤ sup,f(r),rE[a,b]re[a,b]1
1ÊÙ ©¥½nÚ Taylor Ðm §5.1 ¼ê4 §5.2 ¥½n ½n 5.2.1 (Rolle). ¼ê f 3 [a, b] þëY§3 (a, b) S§ f(a) = f(b). K3 ξ ∈ (a, b), ¦ f 0 (ξ) = 0. y². ëY¼ê f 34«m [a, b] þ± M Ú m. XJ M = m, K f ð~ê§l f 0 ≡ 0; XJ M > m, Kd f(a) = f(b) m M ¥k´ f 3S: ξ ∈ (a, b) ?¤§d Fermat ½n§f 0 (ξ) = 0. ½n 5.2.2 (Lagrange). ¼ê f 3 [a, b] þëY, 3 (a, b) S, K3 ξ ∈ (a, b), ¦ f 0 (ξ) = f(b) − f(a) b − a , ½ f(b) − f(a) = f 0 (ξ)(b − a). y². - F(x) = f(x) − [f(a) + f(b) − f(a) b − a (x − a)] K F(a) = F(b) = 0, F ÷v½n 1 ^. l ∃ ξ ∈ (a, b), ¦ F 0 (ξ) = 0. d ξ =÷v½n¦ ξ. ½n 5.2.3 (Cauchy). ¼ê f, g 3 [a, b] þëY, 3 (a, b) S, g 0 (x) 6= 0, ∀ x ∈ (a, b). K3 ξ ∈ (a, b), ¦ f(b) − f(a) g(b) − g(a) = f 0 (ξ) g 0(ξ) . y². d½n 1 Ú g 0 6= 0 g(b) 6= g(a). - F(x) = f(x) − [f(a) + f(b) − f(a) g(b) − g(a) (g(x) − g(a))] K F(a) = F(b) = 0, F ÷v½n 1 ^, l ∃ ξ ∈ (a, b), ¦ F 0 (ξ) = 0. ξ = ÷v¦:. 5: - g(x) = x, K Cauchy ½n±íÑ Lagrange ½n. ±þn½nþ¡ ©¥½n. ½n 5.2.4 (È©1¥½n). f, g 3 [a, b] þ Riemann È, g(x) ØCÒ, K ∃ µ, inf x∈[a,b] f(x) ≤ µ ≤ sup x∈[a,b] f(x), 1
2第五章微分中值定理和Taylor展开使得f(a)g(a)dz =g(a)dr证明.不失一般性，可设g(a)≥0.则inf f-g()≤f()g()≤supf-g(r)→inf f:g(a)d≤/f(a)g(r)da≤ sup f :g(a)da上式说明，如果Jg(a)da=0,则Jf(a)g(a)da=0,此时定理当然成立如果g(z)d>0,则令" f(a)g(a)drJ g(r)da则显然有inf fμ≤supf.定理得证口注： (1)当g()=1时，J f()=μ (b-a)(2)特别地，如果f连续，则由介值定理，日E[a,b]，使得f(s)=μ,此时f(r)g(a)da = f(s). / g(n)de.定理5.2.5（积分第二中值定理)．设于在[a,b]上Riemann可积(1)如果g在[a,上单调减，且g(a)≥0,VrE[a,b],则3E[a,b],使得f(r)g()dz = g(a).f(r)dr;(2)如果g在[a,b] 上单调增，且 g(α)≥0, V E[a,b],则日n E[a,bl],使得 (n)g(n)dr = g(6) / (n)dr;(3)一般地，如果g在[a,b]上单调函数，则日（E[a,l]，使得[~ f(n)g(n)da = g(a) - / f()d + g() /~ f(n)dr
2 1ÊÙ ©¥½nÚ Taylor Ðm ¦ Z b a f(x)g(x)dx = µ Z b a g(x)dx. y². Ø5, g(x) ≥ 0. K inf f · g(x) ≤ f(x)g(x) ≤ sup f · g(x) =⇒ inf f · Z b a g(x)dx ≤ Z b a f(x)g(x)dx ≤ sup f · Z b a g(x)dx þª`², XJ R b a g(x)dx = 0, K R b a f(x)g(x)dx = 0, d½n,¤á. XJ R b a g(x)dx > 0, K- µ = R b a f(x)g(x)dx R b a g(x)dx Kw,k inf f ≤ µ ≤ sup f. ½ny. 5: (1) g(x) ≡ 1 , R b a f(x) = µ · (b − a). (2) AO/, XJ f ëY, Kd0½n, ∃ ξ ∈ [a, b], ¦ f(ξ) = µ, d Z b a f(x)g(x)dx = f(ξ) · Z b a g(x)dx. ½n 5.2.5 (È©1¥½n). f 3 [a, b] þ Riemann È. (1) XJ g 3 [a, b] þüN~, g(x) ≥ 0, ∀ x ∈ [a, b], K ∃ ξ ∈ [a, b], ¦ Z b a f(x)g(x)dx = g(a) · Z ξ a f(x)dx; (2) XJ g 3 [a, b] þüNO, g(x) ≥ 0, ∀ x ∈ [a, b], K ∃ η ∈ [a, b], ¦ Z b a f(x)g(x)dx = g(b) · Z b η f(x)dx; (3) /, XJ g 3 [a, b] þüN¼ê, K ∃ ζ ∈ [a, b], ¦ Z b a f(x)g(x)dx = g(a) · Z ζ a f(x)dx + g(b) · Z b ζ f(x)dx.
3$5.3L'Hospital法则证明.我们只证明一个特殊情形：于连续，9连续可微，9’≤0.令F(）=f(t)dt,则F"=f(a).从而f(a)g()da =F (r)g(r)daFg'da= F(r)g(r)临 -= F(b) · g(b) - μ-g'da= F(b) · g(b) - μ(g(6) - g(a))这说明inf F-g(a)≤f(r)g(r)dr≤supF.g(a)因此,存在[a,b],使得f(a)g(a)dt = F()g(a) = g(a) - [f(r)drg5.3L'Hospital法则Motivation：设f.g为函数，求极限f(a)lim+10 g(r)困难的情形：(1) →o 时, f() →0, g() →0,(%) 型;(2)→0时, f()→80, g()→80,() 型定理5.3.1(L'Hospital)：设fg在(a,b)内可导，且g(a)≠0,VE(a,b).又设lim f(r) = 0 = lim g(z)如果极限f'(α)lim+a+ g'(r)存在(或为00),则f'(a)f(a)limlim+a+ g(r)r→a+ g'(r)证明.补充定义f(a)=g(a)=0,则于在[a,b)上连续.由 Cauchy中值定理V(a,b),(a,),使得f(a)f(r) - f(a)f'(6)g()g(z)g() - g(a)
§5.3 L’Hospital {K 3 y². ·y²AÏ/: f ëY, g ëY, g 0 ≤ 0. - F(x) = R x a f(t)dt, K F 0 = f(x). l Z b a f(x)g(x)dx = Z b a F 0 (x)g(x)dx = F(x)g(x)| b a − Z b a F g0 dx = F(b) · g(b) − µ · Z b a g 0 dx = F(b) · g(b) − µ(g(b) − g(a)). ù`² inf F · g(a) ≤ Z b a f(x)g(x)dx ≤ supF · g(a). Ïd, 3 ξ ∈ [a, b], ¦ Z b a f(x)g(x)dx = F(ξ)g(a) = g(a) · Z ξ a f(x)dx. §5.3 L’Hospital {K Motivation: f, g ¼ê, ¦4 limx→x0 f(x) g(x) (J/: (1) x → x0 , f(x) → 0, g(x) → 0, ( 0 0 ) .; (2) x → x0 , f(x) → ∞, g(x) → ∞, (∞ ∞) . ½n 5.3.1 (L’Hospital). f, g 3 (a, b) S, g(x) 6= 0, ∀ x ∈ (a, b). q lim x→a+ f(x) = 0 = lim x→a+ g(x) XJ4 lim x→a+ f 0 (x) g 0(x) 3 (½ ∞), K lim x→a+ f(x) g(x) = lim x→a+ f 0 (x) g 0(x) y². Ö¿½Â f(a) = g(a) = 0, K f 3 [a, b) þëY. d Cauchy ¥½n, ∀ x ∈ (a, b), ∃ ξ ∈ (a, x), ¦ f(x) g(x) = f(x) − f(a) g(x) − g(a) = f 0 (ξ) g 0(ξ) .
4第五章微分中值定理和Taylor展开当E-→a+时.→a+.从而f()→ limf'(a)g'(s)T-a+ g'()因此f(r)f'(r)limlimi+ g'(α)a+g()注：(1)如果仍有f(a+）=g(α+）=0,则可利用二次导数继续求：f(μ)f'(r)f"(r)limlimlimra+ g(a)=→α+ g (a)→a+ g"(r)高阶导数的情形类似(2)区间(a,b)换成(-0,b)或(a,80)时，有类似结论f(α)f'(a)limlim-00 g()0g()-f(r)f'(r)limlimF00g()+o0 g'(r)（作变量代换工=1即可）定理5.3.2(L'Hospital).设f,g在(a,b）内可导，g(r)≠0且lim g(a) = 00,-0如果极限f(r)lima+ g()存在（或为80),则f'(r)f(r)= limlimr-a+ g(r)r-a+ g'(μ)证明.我们对f'(a)I = lim2<8~a+ g(r)的情形加以证明,1=8的情形类似的可证.>0,>0,使得当(a,a+)时1-=<f(a)<I+E.g'(α)取c=a+，则由Cauchy中值公式，日E（a,c)，使得f'()f(r) f(a)1-<<l+e, rE(a,c)g()g(r) - g(a)
4 1ÊÙ ©¥½nÚ Taylor Ðm x → a + , ξ → a +, l f 0 (ξ) g 0(ξ) → lim x→a+ f 0 (x) g 0(x) , Ïd lim x→a+ f(x) g(x) = lim x→a+ f 0 (x) g 0(x) . 5: (1) XJEk f 0 (a +) = g 0 (a +) = 0, K|^gêUY¦: lim x→a+ f(x) g(x) = lim x→a+ f 0 (x) g 0(x) = lim x→a+ f 00(x) g 00(x) , p ê/aq. (2) «m (a, b) ¤ (−∞, b) ½ (a, ∞) , kaq(Ø: lim x→−∞ f(x) g(x) = lim x→−∞ f 0 (x) g 0(x) , lim x→+∞ f(x) g(x) = lim x→+∞ f 0 (x) g 0(x) . (Cþ x = 1 t =). ½n 5.3.2 (L’Hospital). f, g 3 (a, b) S, g(x) 6= 0 lim x→a+ g(x) = ∞, XJ4 lim x→a+ f(x) g(x) 3 ( ½ ∞), K lim x→a+ f(x) g(x) = lim x→a+ f 0 (x) g 0(x) . y². ·é l = lim x→a+ f 0 (x) g 0(x) < ∞ /\±y², l = ∞ /aqy. ∀ ε > 0, ∃ δ > 0, ¦ x ∈ (a, a+δ) l − ε < f 0 (x) g 0(x) < l + ε. c = a + δ 2 , Kd Cauchy ¥úª, ∃ ξ ∈ (a, c), ¦ l − ε < f(x) − f(a) g(x) − g(a) = f 0 (ξ) g 0(ξ) < l + ε, x ∈ (a, c).
5$5.3L'Hospital法则因为→a+时，g()→80,故在上式中令→a，得f'(r)1-es lim inf f<limsup<l+eg(r)g(r)-T0因为是任意取的，令→0,就得f'(a)lim inf f'()limsup=1g(r)g(r)t→a+z-a+注和定理1的注记一样，有类似的注记（略）例5.3.1.求极限r-sinlimr3T-0解sina11-cosra-sinr= limlim= lim6T33r22006元20例5.3.2.设f"(ro）存在，求极限f(ro +h)-f(zo)-f(ro)hlim :h2h-0解f(to+h)-f(ro)-f(ro)hh = lim (c + h) - (c) limf"(ro)h22hh-0h_0例5.3.3.设1E(0,1),an+1=n(1-n),n=1,2,...证明n.n→1.证明易见0<an<1,Vn≥1.从而Tn+1= Tn·(1 -In)<an设n→a,则在上式中令n→8,得a=a.(l-a)→a=0.即m→0.从而111lim (-)= lim=1001-InEnn=00n+1-因此1111(lim-)= 1lim1+nTn1n-0onann1-k=1这说明lim n.an =1
§5.3 L’Hospital {K 5 Ï x → a+ , g(x) → ∞, 3þª¥- x → a +, l − ε ≤ lim x→a+ inf f 0 (x) g 0(x) ≤ lim x→a+ sup f 0 (x) g 0 (x) ≤ l + ε. Ï ε ´?¿, - ε → 0, Ò lim x→a+ inf f 0 (x) g 0(x) = lim x→a+ sup f 0 (x) g 0(x) = l. 5 Ú½n 1 5P, kaq5P (Ñ). ~ 5.3.1. ¦4 limx→0 x − sin x x 3 . ) lim x→0 x − sin x x 3 = lim x→0 1 − cos x 3x 2 = lim x→0 sin x 6x = 1 6 . ~ 5.3.2. f 00(x0) 3, ¦4 lim h→0 f(x0 + h) − f(x0) − f 0 (x0)h h 2 . ) lim h→0 f(x0 + h) − f(x0) − f 0 (x0)h h 2 = lim h→0 f 0 (x0 + h) − f 0 (x0) 2h = 1 2 f 00(x0). ~ 5.3.3. x1 ∈ (0, 1), xn+1 = xn · (1 − xn), n = 1, 2, · · ·. y² n · xn → 1. y² ´ 0 < xn < 1, ∀ n ≥ 1. l xn+1 = xn · (1 − xn) < xn. xn → a, K3þª¥- n → ∞, a = a · (1 − a) ⇒ a = 0. = xn → 0. l limn→∞ ( 1 xn+1 − 1 xn ) = limn→∞ 1 1 − xn = 1 Ïd limn→∞ 1 n · ( 1 xn − 1 x1 ) = limn→∞ 1 n · nX−1 k=1 ( 1 xn − 1 xn ) = 1, ù`² limn→∞ n · xn = 1.