6 (3)limv(y)=c存在 证明 (i)累次极限 lim lim f(x,y)=limo(x)=c存在; (i)累次极限 limlim f(x,y)=limv(y)=c存在; (i)二重极限limf(x,y)=c 44、设函数u=∫(x,y,2)在闭立方体a≤x≤b,a≤y≤b,a≤z≤b上连续,令叭x) mmf(x,y,2)试证明。φ在上连续 45、若函数f(x,y)在矩形D:-a≤x≤a,-b≤y≤b(a>0,b>0)上分别为x和y的连续函 数,而且f(0,0)=0。当x固定时,f(x,y)是y的严格递减函数,则有6>0,使对每个x∈(-6,6) 有y∈(-b,b)满足f(x,y)=0 Hint: Because the function f(0, y) is strictly decreasing on the interval -b, b, and f(0, 0)=0, then f(0, b)<0, f(0, -b)>0. Further, consider to use the continuity of the function f 46、若f(x,y)函数在xy平面上处处连续,试证平面点集E={(x,y)f(x,y)=0}为闭集 47、设∫在P上连续,x≠O(中的原点)时,f(x)>0,且x∈Rn,及c>0有f(cx)=cf(x) 试证明:彐a,b>0,使得a≤f(x)≤b,(vx∈R) 48、设A是n×n矩阵,detA≠0.试证明:a>0,B>0使得∈R有x≤|Aarl≤a|r R,满足如下三条件: (1)vx∈Rn,f(x)≥0,且x=O←→f(x)=0: (2)VA∈R,有f(x)=|f(x); (3)vr,y∈Rn,f(x+y)≤f(x)+f(y) 试证明:()M>0,使得Vx∈R",有f(x)≤M|c (i)f满足 Lipschitz条件:即>0,使得|f(x)-f(列≤Lx-y(x,y∈R); (i)常数a>0使得∈R有ar≤f(x) 50、设f在Rn中点x0的邻域里有界,记Mf(xo,6)=sup{f()x∈Rn,|-xol<b} (x0,6)=inf{f(x)x∈P",|x-ol<}.则(1)极限r(xo)=:lim[M(xo,0)-mr(xo,6)存 在,并称之为∫在x处的振幅;(2)f(x)在xo连续的充分必要条件是ur(xo)=0 51、设集合ECRn是非闭的,函数f在E上一致连续。试证明:f只能惟一地连续延拓到E 上,即惟一地存在E上的函数F满足F(x)=f(x),x∈E,使得F在E上一致连续 52、设∫在R上连续,=m、f)存在。试证明f在P上一致连续6 (3) limy→b ψ(y) = c   (i) tu" limx→a lim y→b f(x, y) = limx→a φ(x) = c 5 (ii) tu" lim y→b limx→a f(x, y) = limy→b ψ(y) = c 5 (iii) Gh" lim (x,y)→(a,b) f(x, y) = c. 44  u = f(x, y, z) Z@vM a ≤ x ≤ b, a ≤ y ≤ b, a ≤ z ≤ b 074. φ(x) = max a≤y≤b { min a≤z≤b f(x, y, z)}. 3 φ  [a, b] 074 45 ( f(x, y) wi D : −a ≤ x ≤ a, −b ≤ y ≤ b (a > 0,b > 0) 0Ex x  y 74 y& f(0, 0) = 0 : x z6 f(x, y) ! y &'( δ > 0 >;< x ∈ (−δ, δ)  y ∈ (−b, b) 2+ f(x, y)=0. Hint: Because the function f(0, y) is strictly decreasing on the interval [−b, b], and f(0, 0) = 0, then f(0, b) < 0, f(0, −b) > 0. Further, consider to use the continuity of the function f. 46 ( f(x, y)  xy _m0MM743_mJ[ E = {(x, y)|f(x, y)=0} Z[ 47 f  Rn 074x = O(Rn 8jJ) 6f(x) > 0, & ∀x ∈ Rn, / c > 0  f(cx) = cf(x). 3 ∃a, b > 0,  a|x| ≤ f(x) ≤ b|x|,(∀x ∈ Rn). 48 A ! n × n wk detA = 0. 3 ∃α > 0,β > 0  ∀x ∈ Rn  β|x|≤|Ax| ≤ α|x|. 49∗ f : Rn → R, 2+$];?F (1) ∀x ∈ Rn, f(x) ≥ 0, & x = O ⇐⇒ f(x) = 0; (2) ∀λ ∈ R,  f(λx) = |λ|f(x); (3) ∀x, y ∈ Rn, f(x + y) ≤ f(x) + f(y). 3 (i) ∃M > 0,  ∀x ∈ Rn,  f(x) ≤ M|x|; (ii) f 2+ Lipschitz ?F_ ∃L > 0,  |f(x) − f(y)| ≤ L|x − y|(∀x, y ∈ Rn); (iii) ∃ 1 α > 0  ∀x ∈ Rn  α|x| ≤ f(x). 50 f  Rn 8J x0 pO{4| Mf (x0, δ) = sup{f(x)   x ∈ Rn, |x − x0| < δ}, mf (x0, δ) = inf{f(x)   x ∈ Rn, |x − x0| < δ}  1 " ωf (x0) =: limδ→0+  Mf (x0, δ) − mf (x0, δ)   6i< f  x0 Ml}5 2  f(x)  x0 74DE!>?F! ωf (x0)=0. 51 [\ E ⊂ Rn !~Z f  E 09a743 f mno974pqq E 0_o9 E 0 F 2+ F(x) = f(x), x ∈ E,  F  E 09a74 52 f  Rn 074 lim |x|→+∞ f(x) 3 f  Rn 09a74