得分评卷人 三、证明题（本大题共5小题，每题6分，共30分） 1。证明下列方程在指定点的邻域存在以工为自变量的隐函数，并求警：割= xe+1,点(0,1). 证明：令F(x,)=y-xe-1,则F(0,1)=1-0×e-1=0 F 盟=+w鼎→1-2=心→是- 2.证明广嘉证收敛 证明：因为∈1，+o0)有品≤m而 一乐甲 =V1-1<+o,A->1. 所以由比较判剥法山收敛。再用比较判别法高血收敛 3.证明6冯d止收敛 证明：x=1是汽的暇点， 卿1-到@ (1-x)x3 气-卿a-+2 -即1+i0+网27-+e,A=2<1 所以由比较判别法6温收敛. 数学分析四试题第6页（共8页）✚➞ ➭ò❁ ♥✦②➨❑ (✢➀❑✁ 5 ✂❑➜③❑ 6 ➞➜✁ 30 ➞ ) 1. ②➨❡✎➄➜✸➁➼✿✛✙➁⑧✸➧ x ➃❣❈þ✛Û➻ê, ➾➛ dy dx : y = xey + 1, ✿ (0, 1). ②➨: ✲ F(x, y) = y − xey − 1, ❑ F(0, 1) = 1 − 0 × e 1 − 1 = 0; ∂F ∂x = −e y , ∂F ∂y = 1 − xey ✸ R2 ë❨; ∂F ∂y |(0,1) = 1 − 0 × e 1 = 1 6= 0. ↕➧❞Û➻ê⑧✸➼♥ ∃δ > 0, ∀x ∈ (−δ, δ) ⑧✸➁➌➌❻❦ë❨✓ê✛Û➻ê y = f(x) ➛✚ F[x, f(x)] ≡ 0, f(0) = 1. é y = xey + 1 ✛ü❃➛✬✉ x ✛✓ê❦ dy dx = e y + xey dy dx ⇒ (1 − xey ) dy dx = e y ⇒ dy dx = e y 1 − xey . 2. ②➨ R +∞ 1 cos x x √ x+1 dx ➶ñ. ②➨: Ï➃ ∀x ∈ [1, +∞) ❦    cos x x √ x+1    ≤ 1 x √ x+1 . ✌ lim x→+∞ x 3 2 1 x √ x + 1 = lim x→+∞ r x x + 1 = r lim x→+∞ x x + 1 = √ 1 = 1 < +∞, λ = 3 2 > 1. ↕➧❞✬✖✞❖④ R +∞ 1 1 x √ x+1 dx ➶ñ. ✷❫✬✖✞❖④ R +∞ 1 cos x x √ x+1 dx ➶ñ. 3. ②➨ R 1 0 √ x √ 1−x4 dx ➶ñ. ②➨: x = 1 ➫ √ x √ 1−x4 ✛❛✿. limx→1− (1 − x) 1 2 √ x √ 1 − x 4 = limx→1− (1 − x) 1 2 x 1 2 (1 − x 2 ) 1 2 (1 + x 2 ) 1 2 = limx→1− x 1 2 (1 + x) 1 2 (1 + x 2 ) 1 2 = 1 √ 2 √ 2 = 1 2 < +∞, λ = 1 2 < 1. ↕➧❞✬✖✞❖④ R 1 0 √ x √ 1−x4 dx ➶ñ. ê➷➞Û(III)➪❑ ✶ 6 ➄↔✁ 8 ➄↕