高等数学（第七版）同济大学习题9-1 个人解答

高等数学（第七版）同济大学习题9-1

1.判定下列平面点集中哪些是开集、闭集、区域、有界集、无界集？并分别指出它们的聚点所成的点集（称为导集）和边界.\begin{aligned}&1. \ 判定下列平面点集中哪些是开集、闭集、区域、有界集、无界集？并分别指出它们的聚点所成的点集（称为导集）和边界.&\end{aligned}1. 判定下列平面点集中哪些是开集、闭集、区域、有界集、无界集？并分别指出它们的聚点所成的点集（称为导集）和边界.

(1){(x,y)∣x≠0,y≠0}； (2){(x,y)∣1<x2+y2≤4}；(3){(x,y)∣y>x2}；(4){(x,y)∣x2+(y−1)2≥1}∩{(x,y)∣x2+(y−2)2≤4}.\begin{aligned} &\ \ (1)\ \ \{(x, \ y)\ |\ x \neq 0, \ y \neq 0\}；\ \ \ \ \ \ \ \ \ \ \ (2)\ \ \{(x, \ y)\ |\ 1 \lt x^2+y^2 \le 4\}；\\\\ &\ \ (3)\ \ \{(x, \ y)\ |\ y \gt x^2\}；\\\\ &\ \ (4)\ \ \{(x, \ y)\ |\ x^2+(y-1)^2 \ge 1\} \cap \{(x, \ y)\ |\ x^2+(y-2)^2 \le 4\}. & \end{aligned} (1) {(x, y) ∣ x=0, y=0}； (2) {(x, y) ∣ 1<x2+y2≤4}； (3) {(x, y) ∣ y>x2}； (4) {(x, y) ∣ x2+(y−1)2≥1}∩{(x, y) ∣ x2+(y−2)2≤4}.

解：

(1)集合是开集，无界集，导集为R2，边界为{(x,y)∣x=0或y=0}.(2)集合既非开集，又非闭集，是有界集，导集为{(x,y)∣1≤x2+y2≤4}，边界为{(x,y)∣x2+y2=1}∪{(x,y)∣x2+y2=4}.(3)集合是开集，区域，无界集，导集为{(x,y)∣y≥x2}，边界为{(x,y)∣y=x2}.(4)集合是闭集，有界集，导集是集合本身，边界为{(x,y)∣x2+(y−1)2=1}∪{(x,y)∣x2+(y−2)2=4}.\begin{aligned} &\ \ (1)\ 集合是开集，无界集，导集为R^2，边界为\{(x, \ y)\ |\ x=0或y=0\}.\\\\ &\ \ (2)\ 集合既非开集，又非闭集，是有界集，导集为\{(x, \ y)\ |\ 1\le x^2+y^2 \le 4\}，\\\\ &\ \ \ \ \ \ \ \ 边界为\{(x,\ y)\ |\ x^2+y^2=1\} \cup \{(x, \ y)\ |\ x^2+y^2=4\}.\\\\ &\ \ (3)\ 集合是开集，区域，无界集，导集为\{(x, \ y)\ |\ y \ge x^2\}，边界为\{(x, \ y)\ |\ y=x^2\}.\\\\ &\ \ (4)\ 集合是闭集，有界集，导集是集合本身，边界为\{(x, \ y)\ |\ x^2+(y-1)^2=1\} \cup \{(x,\ y)\ |\ x^2+(y-2)^2=4 \}. & \end{aligned} (1) 集合是开集，无界集，导集为R2，边界为{(x, y) ∣ x=0或y=0}. (2) 集合既非开集，又非闭集，是有界集，导集为{(x, y) ∣ 1≤x2+y2≤4}，边界为{(x, y) ∣ x2+y2=1}∪{(x, y) ∣ x2+y2=4}. (3) 集合是开集，区域，无界集，导集为{(x, y) ∣ y≥x2}，边界为{(x, y) ∣ y=x2}. (4) 集合是闭集，有界集，导集是集合本身，边界为{(x, y) ∣ x2+(y−1)2=1}∪{(x, y) ∣ x2+(y−2)2=4}.

2.已知函数f(x,y)=x2+y2−xytanxy，试求f(tx,ty).\begin{aligned}&2. \ 已知函数f(x, \ y)=x^2+y^2-xytan\ \frac{x}{y}，试求f(tx, \ ty).&\end{aligned}2. 已知函数f(x, y)=x2+y2−xytan yx，试求f(tx, ty).

解：

f(tx,ty)=(tx)2+(ty)2−(tx)(ty)tantxty=t2(x2+y2−xytanxy)=t2f(x,y)\begin{aligned} &\ \ f(tx, \ ty)=(tx)^2+(ty)^2-(tx)(ty)tan\ \frac{tx}{ty}=t^2\left(x^2+y^2-xytan\ \frac{x}{y}\right)=t^2f(x, \ y) & \end{aligned} f(tx, ty)=(tx)2+(ty)2−(tx)(ty)tan tytx=t2(x2+y2−xytan yx)=t2f(x, y)

3.试证函数F(x,y)=lnx⋅ln⁡y满足关系式F(xy,uv)=F(x,u)+F(x,v)+F(y,u)+F(y,v).\begin{aligned}&3. \ 试证函数F(x,\ y)=ln\ x\cdot \ln\ y满足关系式F(xy, \ uv)=F(x, \ u)+F(x, \ v)+F(y, \ u)+F(y, \ v).&\end{aligned}3. 试证函数F(x, y)=ln x⋅ln y满足关系式F(xy, uv)=F(x, u)+F(x, v)+F(y, u)+F(y, v).

解：

F(xy,uv)=ln(xy)⋅ln(uv)=(lnx+lny)(lnu+lnv)=lnx⋅lnu+lnx⋅lnv+lny⋅lnu+lny⋅lnv=F(x,u)+F(x,v)+F(y,u)+F(y,v)\begin{aligned} &\ \ F(xy, \ uv)=ln(xy)\cdot ln(uv)=(ln\ x+ln\ y)(ln\ u+ln\ v)=ln\ x\cdot ln\ u+ln\ x\cdot ln\ v+ln\ y\cdot ln\ u+ln\ y\cdot ln\ v=\\\\ &\ \ F(x, \ u)+F(x, \ v)+F(y, \ u)+F(y, \ v) & \end{aligned} F(xy, uv)=ln(xy)⋅ln(uv)=(ln x+ln y)(ln u+ln v)=ln x⋅ln u+ln x⋅ln v+ln y⋅ln u+ln y⋅ln v= F(x, u)+F(x, v)+F(y, u)+F(y, v)

4.已知函数f(u,v,w)=uw+wu+v，试求f(x+y,x−y,xy).\begin{aligned}&4. \ 已知函数f(u, \ v, \ w)=u^w+w^{u+v}，试求f(x+y, \ x-y, \ xy).&\end{aligned}4. 已知函数f(u, v, w)=uw+wu+v，试求f(x+y, x−y, xy).

解：

f(x+y,x−y,xy)=(x+y)xy+(xy)(x+y)+(x−y)=(x+y)xy+(xy)2x\begin{aligned} &\ \ f(x+y, \ x-y, \ xy)=(x+y)^{xy}+(xy)^{(x+y)+(x-y)}=(x+y)^{xy}+(xy)^{2x} & \end{aligned} f(x+y, x−y, xy)=(x+y)xy+(xy)(x+y)+(x−y)=(x+y)xy+(xy)2x

5.求下列各函数的定义域：\begin{aligned}&5. \ 求下列各函数的定义域：&\end{aligned}5. 求下列各函数的定义域：

(1)z=ln(y2−2x+1)； (2)z=1x+y+1x−y；(3)z=x−y； (4)z=ln(y−x)+x1−x2−y2；(5)u=R2−x2−y2−z2+1x2+y2+z2−r2(R>r>0)；(6)u=arccoszx2+y2.\begin{aligned} &\ \ (1)\ \ z=ln(y^2-2x+1)；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ z=\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{x-y}}；\\\\ &\ \ (3)\ \ z=\sqrt{x-\sqrt{y}}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ z=ln(y-x)+\frac{\sqrt{x}}{\sqrt{1-x^2-y^2}}；\\\\ &\ \ (5)\ \ u=\sqrt{R^2-x^2-y^2-z^2}+\frac{1}{\sqrt{x^2+y^2+z^2-r^2}}\ (R \gt r \gt 0)；\\\\ &\ \ (6)\ \ u=arccos\ \frac{z}{x^2+y^2}. & \end{aligned} (1) z=ln(y2−2x+1)； (2) z=x+y1+x−y1； (3) z=x−y； (4) z=ln(y−x)+1−x2−y2x； (5) u=R2−x2−y2−z2+x2+y2+z2−r21 (R>r>0)； (6) u=arccos x2+y2z.

解：

(1){(x,y)∣y2−2x+1>0}(2){(x,y)∣x+y>0,x−y>0}(3){(x,y)∣x≥0,y≥0,x2≥y}(4){(x,y)∣y−x>0,x≥0,x2+y2<1}(5){(x,y,z)∣r2<x2+y2+z2≤R2}(6){(x,y,z)∣x2+y2−z2≥0,x2+y2≠0}\begin{aligned} &\ \ (1)\ \{(x, \ y)\ |\ y^2-2x+1 \gt 0\}\\\\ &\ \ (2)\ \{(x,\ y)\ |\ x+y \gt 0,\ x-y \gt 0\}\\\\ &\ \ (3)\ \{(x, \ y)\ |\ x \ge 0, \ y \ge 0, \ x^2 \ge y\}\\\\ &\ \ (4)\ \{(x, \ y)\ |\ y-x \gt 0, \ x \ge 0, \ x^2+y^2 \lt 1\}\\\\ &\ \ (5)\ \{(x, \ y, \ z)\ |\ r^2 \lt x^2+y^2+z^2 \le R^2\}\\\\ &\ \ (6)\ \{(x, \ y, \ z)\ |\ x^2+y^2-z^2 \ge 0, \ x^2+y^2 \neq 0\} & \end{aligned} (1) {(x, y) ∣ y2−2x+1>0} (2) {(x, y) ∣ x+y>0, x−y>0} (3) {(x, y) ∣ x≥0, y≥0, x2≥y} (4) {(x, y) ∣ y−x>0, x≥0, x2+y2<1} (5) {(x, y, z) ∣ r2<x2+y2+z2≤R2} (6) {(x, y, z) ∣ x2+y2−z2≥0, x2+y2=0}

6.求下列各极限：\begin{aligned}&6. \ 求下列各极限：&\end{aligned}6. 求下列各极限：

(1)lim⁡(x,y)→(0,1)1−xyx2+y2； (2)lim⁡(x,y)→(1,0)ln(x+ey)x2+y2；(3)lim⁡(x,y)→(0,0)2−xy+4xy； (4)lim⁡(x,y)→(0,0)xy2−exy−1；(5)lim⁡(x,y)→(2,0)tan(xy)y； (6)lim⁡(x,y)→(0,0)1−cos(x2+y2)(x2+y2)ex2y2.\begin{aligned} &\ \ (1)\ \ \lim_{(x,\ y)\rightarrow (0, \ 1)}\frac{1-xy}{x^2+y^2}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ \lim_{(x,\ y)\rightarrow (1, \ 0)}\frac{ln(x+e^y)}{\sqrt{x^2+y^2}}；\\\\ &\ \ (3)\ \ \lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{2-\sqrt{xy+4}}{xy}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ \lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{xy}{\sqrt{2-e^{xy}}-1}；\\\\ &\ \ (5)\ \ \lim_{(x,\ y)\rightarrow (2, \ 0)}\frac{tan(xy)}{y}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)\ \ \lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{1-cos(x^2+y^2)}{(x^2+y^2)e^{x^2y^2}}. & \end{aligned} (1) (x, y)→(0, 1)limx2+y21−xy； (2) (x, y)→(1, 0)limx2+y2ln(x+ey)； (3) (x, y)→(0, 0)limxy2−xy+4； (4) (x, y)→(0, 0)lim2−exy−1xy； (5) (x, y)→(2, 0)limytan(xy)； (6) (x, y)→(0, 0)lim(x2+y2)ex2y21−cos(x2+y2).

解：

(1)lim⁡(x,y)→(0,1)1−xyx2+y2=1−00+1=1.(2)lim⁡(x,y)→(1,0)ln(x+ey)x2+y2=ln(1+e0)1+0=ln2.(3)lim⁡(x,y)→(0,0)2−xy+4xy=lim⁡(x,y)→(0,0)(2−xy+4)(2+xy+4)xy(2+xy+4)=lim⁡(x,y)→(0,0)4−(xy+4)xy(2+xy+4)=−12+4=−14.(4)lim⁡(x,y)→(0,0)xy2−exy−1=lim⁡(x,y)→(0,0)xy1−exy⋅(2−exy+1)=−1⋅2=−2.(5)lim⁡(x,y)→(2,0)tan(xy)y=lim⁡(x,y)→(2,0)tan(xy)xy⋅x=1⋅2=2.(6)lim⁡(x,y)→(0,0)1−cos(x2+y2)(x2+y2)ex2y2=lim⁡(x,y)→(0,0)1−cos(x2+y2)(x2+y2)2⋅x2+y2ex2y2=12⋅0=0.\begin{aligned} &\ \ (1)\ \lim_{(x,\ y)\rightarrow (0, \ 1)}\frac{1-xy}{x^2+y^2}=\frac{1-0}{0+1}=1.\\\\ &\ \ (2)\ \lim_{(x,\ y)\rightarrow (1, \ 0)}\frac{ln(x+e^y)}{\sqrt{x^2+y^2}}=\frac{ln(1+e^0)}{\sqrt{1+0}}=ln\ 2.\\\\ &\ \ (3)\ \lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{2-\sqrt{xy+4}}{xy}=\lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{(2-\sqrt{xy+4})(2+\sqrt{xy+4})}{xy(2+\sqrt{xy+4})}=\lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{4-(xy+4)}{xy(2+\sqrt{xy+4})}=\frac{-1}{2+\sqrt{4}}=-\frac{1}{4}.\\\\ &\ \ (4)\ \lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{xy}{\sqrt{2-e^{xy}}-1}=\lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{xy}{1-e^{xy}}\cdot (\sqrt{2-e^{xy}}+1)=-1\cdot 2=-2.\\\\ &\ \ (5)\ \lim_{(x,\ y)\rightarrow (2, \ 0)}\frac{tan(xy)}{y}=\lim_{(x,\ y)\rightarrow (2, \ 0)}\frac{tan(xy)}{xy}\cdot x=1\cdot 2=2.\\\ &\ \ (6)\ \lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{1-cos(x^2+y^2)}{(x^2+y^2)e^{x^2y^2}}=\lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{1-cos(x^2+y^2)}{(x^2+y^2)^2}\cdot \frac{x^2+y^2}{e^{x^2y^2}}=\frac{1}{2}\cdot 0=0. & \end{aligned} (1) (x, y)→(0, 1)limx2+y21−xy=0+11−0=1. (2) (x, y)→(1, 0)limx2+y2ln(x+ey)=1+0ln(1+e0)=ln 2. (3) (x, y)→(0, 0)limxy2−xy+4=(x, y)→(0, 0)limxy(2+xy+4)(2−xy+4)(2+xy+4)=(x, y)→(0, 0)limxy(2+xy+4)4−(xy+4)=2+4−1=−41. (4) (x, y)→(0, 0)lim2−exy−1xy=(x, y)→(0, 0)lim1−exyxy⋅(2−exy+1)=−1⋅2=−2. (5) (x, y)→(2, 0)limytan(xy)=(x, y)→(2, 0)limxytan(xy)⋅x=1⋅2=2. (6) (x, y)→(0, 0)lim(x2+y2)ex2y21−cos(x2+y2)=(x, y)→(0, 0)lim(x2+y2)21−cos(x2+y2)⋅ex2y2x2+y2=21⋅0=0.

7.证明下列极限不存在：\begin{aligned}&7. \ 证明下列极限不存在：&\end{aligned}7. 证明下列极限不存在：

(1)lim⁡(x,y)→(0,0)x+yx−y； (2)lim⁡(x,y)→(0,0)x2y2x2y2+(x−y)2.\begin{aligned} &\ \ (1)\ \ \lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{x+y}{x-y}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ \lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{x^2y^2}{x^2y^2+(x-y)^2}. & \end{aligned} (1) (x, y)→(0, 0)limx−yx+y； (2) (x, y)→(0, 0)limx2y2+(x−y)2x2y2.

解：

(1)当(x,y)沿直线y=kx趋于(0,0)时，有lim⁡(x,y)→(0,0)x+yx−y=lim⁡(x,y)→(0,0)(1+k)x(1−k)x=1+k1−k(k≠1)，极限随k值的不同而改变，所以极限不存在.(2)取(x,y)→(0,0)的两种方式：y=x，y=−x，分别求极限，lim⁡(x,y)→(0,0)x2y2x2y2+(x−y)2=lim⁡x→0x4x4=1，lim⁡(x,y)→(0,0)x2y2x2y2+(x−y)2=lim⁡x→0x4x4+4x2=lim⁡x→0x2x2+4=0，两种方式取得极限不同，所以极限不存在.\begin{aligned} &\ \ (1)\ 当(x, \ y)沿直线y=kx趋于(0, \ 0)时，有\lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{x+y}{x-y}=\lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{(1+k)x}{(1-k)x}=\frac{1+k}{1-k}\ (k \neq 1)，\\\\ &\ \ \ \ \ \ \ \ 极限随k值的不同而改变，所以极限不存在.\\\\ &\ \ (2)\ 取(x, \ y)\rightarrow(0, \ 0)的两种方式：y=x，y=-x，分别求极限，\lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{x\rightarrow 0}\frac{x^4}{x^4}=1，\\\\ &\ \ \ \ \ \ \ \ \lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{x^2y^2}{x^2y^2+(x-y)^2}=\lim_{x \rightarrow 0}\frac{x^4}{x^4+4x^2}=\lim_{x \rightarrow 0}\frac{x^2}{x^2+4}=0，两种方式取得极限不同，所以极限不存在. & \end{aligned} (1) 当(x, y)沿直线y=kx趋于(0, 0)时，有(x, y)→(0, 0)limx−yx+y=(x, y)→(0, 0)lim(1−k)x(1+k)x=1−k1+k (k=1)，极限随k值的不同而改变，所以极限不存在. (2) 取(x, y)→(0, 0)的两种方式：y=x，y=−x，分别求极限，(x, y)→(0, 0)limx2y2+(x−y)2x2y2=x→0limx4x4=1， (x, y)→(0, 0)limx2y2+(x−y)2x2y2=x→0limx4+4x2x4=x→0limx2+4x2=0，两种方式取得极限不同，所以极限不存在.

8.函数z=y2+2xy2−2x在何处是间断的？\begin{aligned}&8. \ 函数z=\frac{y^2+2x}{y^2-2x}在何处是间断的？&\end{aligned}8. 函数z=y2−2xy2+2x在何处是间断的？

解：

函数的定义域为D={(x,y)∣y2−2x≠0}，曲线y2−2x=0上各点均为D的聚点，且函数在这些点处没有定义，因此曲线y−2x=0上各点均为函数的间断点.\begin{aligned} &\ \ 函数的定义域为D=\{(x, \ y)\ |\ y^2-2x \neq 0\}，曲线y^2-2x=0上各点均为D的聚点，且函数在这些点处没有定义，\\\\ &\ \ 因此曲线y^-2x=0上各点均为函数的间断点. & \end{aligned} 函数的定义域为D={(x, y) ∣ y2−2x=0}，曲线y2−2x=0上各点均为D的聚点，且函数在这些点处没有定义，因此曲线y−2x=0上各点均为函数的间断点.

9.证明lim⁡(x,y)→(0,0)xyx2+y2=0.\begin{aligned}&9. \ 证明\lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{xy}{\sqrt{x^2+y^2}}=0.&\end{aligned}9. 证明(x, y)→(0, 0)limx2+y2xy=0.

解：

因为∣xyx2+y2−0∣≤12(x2+y2)x2+y2=12x2+y2，要使∣xyx2+y2−0∣<ε，只要x2+y2<2ε，所以∀ε>0，取δ=2ε，则当0<x2+y2<δ时，就有∣xyx2+y2−0∣<ε成立，即lim⁡(x,y)→(0,0)xyx2+y2=0\begin{aligned} &\ \ 因为\left|\frac{xy}{\sqrt{x^2+y^2}}-0\right| \le \frac{\frac{1}{2}(x^2+y^2)}{\sqrt{x^2+y^2}}=\frac{1}{2}\sqrt{x^2+y^2}，要使\left|\frac{xy}{\sqrt{x^2+y^2}}-0\right| \lt \varepsilon，只要\sqrt{x^2+y^2} \lt 2\varepsilon，\\\\ &\ \ 所以 \forall \varepsilon \gt 0，取\delta =2\varepsilon，则当0 \lt \sqrt{x^2+y^2} \lt \delta时，就有\left|\frac{xy}{\sqrt{x^2+y^2}}-0\right| \lt \varepsilon成立，即\lim_{(x,\ y)\rightarrow (0, \ 0)}\frac{xy}{\sqrt{x^2+y^2}}=0 & \end{aligned} 因为∣∣x2+y2xy−0∣∣≤x2+y221(x2+y2)=21x2+y2，要使∣∣x2+y2xy−0∣∣<ε，只要x2+y2<2ε，所以∀ε>0，取δ=2ε，则当0<x2+y2<δ时，就有∣∣x2+y2xy−0∣∣<ε成立，即(x, y)→(0, 0)limx2+y2xy=0

10.设F(x,y)=f(x)，f(x)在x0处连续，证明：对任意y0∈R，F(x,y)在(x0,y0)处连续.\begin{aligned}&10. \ 设F(x, \ y)=f(x)，f(x)在x_0处连续，证明：对任意y_0 \in R，F(x,\ y)在(x_0, \ y_0)处连续.&\end{aligned}10. 设F(x, y)=f(x)，f(x)在x0处连续，证明：对任意y0∈R，F(x, y)在(x0, y0)处连续.

解：

设P0(x0,y0)∈R2，因为f(x)在x0处连续，所以∀ε>0，∃δ>0，当∣x−x0∣<δ时，有∣f(x)−f(x0)∣<ε，从而，当P(x,y)∈U(P0,δ)时，∣x−x0∣≤ρ(P,P0)<δ，因此有∣F(x,y)−F(x0,y0)∣=∣f(x)−f(x0)∣<ε，即F(x,y)在(x0,y0)处连续.\begin{aligned} &\ \ 设P_0(x_0, \ y_0) \in R^2，因为f(x)在x_0处连续，所以\forall \varepsilon \gt 0，\exist \delta \gt 0，当|x-x_0| \lt \delta时，有|f(x)-f(x_0)| \lt \varepsilon，从而，\\\\ &\ \ 当P(x,\ y) \in U(P_0, \ \delta)时，|x-x_0| \le \rho(P, \ P_0) \lt \delta，因此有|F(x, \ y)-F(x_0, \ y_0)|=|f(x)-f(x_0)| \lt \varepsilon，\\\\ &\ \ 即F(x,\ y)在(x_0, \ y_0)处连续. & \end{aligned} 设P0(x0, y0)∈R2，因为f(x)在x0处连续，所以∀ε>0，∃δ>0，当∣x−x0∣<δ时，有∣f(x)−f(x0)∣<ε，从而，当P(x, y)∈U(P0, δ)时，∣x−x0∣≤ρ(P, P0)<δ，因此有∣F(x, y)−F(x0, y0)∣=∣f(x)−f(x0)∣<ε，即F(x, y)在(x0, y0)处连续.