# 微积分代写Calculus III|MATH 15300 University of Chicago Assignment

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The Department of Mathematics offers a range of programs in mathematics and applied mathematics at both undergraduate and graduate levels. The department provides a comprehensive instruction in various areas of mathematics, including pure mathematics, applied mathematics, and statistics.

The undergraduate programs include both a Bachelor of Arts (BA) and a Bachelor of Science (BS) degree in mathematics. The BS program offers two different tracks, one in applied mathematics and one in mathematics with a specialization in economics. The BA program offers a broad foundation in mathematics along with a wide range of electives, allowing students to tailor their program to their specific interests.

In addition to the major programs, students from other fields of study can complete a minor in mathematics. The minor requires a minimum of 21 credit hours in mathematics courses and can be tailored to complement students’ other areas of study.

At the graduate level, the department offers a Master of Science (MS) degree in mathematics and a PhD in mathematics. The graduate programs focus on developing advanced knowledge and research skills in mathematics, and provide opportunities for students to engage in cutting-edge research with faculty members.

Overall, the Department of Mathematics provides a strong academic environment for students interested in mathematics and its applications, with a range of programs at both the undergraduate and graduate levels, as well as opportunities for research and specialization.

Let $f(x, y)=\frac{x^2 y^2}{x^2 y^2+(x+y)^4}$ if $(x, y) \neq(0,0)$. Then $\lim _{(x, y) \rightarrow(0,0)} f(x, y)$ does not exist.

By the remarks preceding this exercise, we need only find two distinct values which the function $f$ assumes in every neighborhood of the origin. This is easy. Every neighborhood of the origin contains points $(x, 0)$ distinct from $(0,0)$ on the $x$-axis. At every such point $f(x, y)=f(x, 0)=0$. Also, every neighborhood of the origin contains points $(x, x)$ distinct from $(0,0)$ which lie on the line $y=x$. At each such point $f(x, y)=f(x, x)=1 / 17$. Thus $f$ has no limit at the origin since in every neighborhood of $(0,0)$ it assumes both the values 0 and $1 / 17$. (Notice, incidentally, that both iterated limits, $\lim {x \rightarrow 0}\left(\lim {y \rightarrow 0} f(x, y)\right)$ and $\lim {y \rightarrow 0}\left(\lim {x \rightarrow 0} f(x, y)\right)$ exist and equal 0.$)$

Suppose $f$ is the constant function defined on $[0,1]$ whose value is 1 . Asked to describe those functions in $\mathcal{B}([0,1])$ which lie in the open ball about $f$ of radius 1 , a student replies (somewhat incautiously) that $B_1(f)$ is the set of all real-valued functions $g$ on $[0,1]$ satisfying $0<g(x)<2$ for all $x \in[0,1]$. Why is this response wrong? (Solution Q.13.1.)

Let $C$ be the set of all functions defined on $[0,1]$ such that $0<g(x)<2$ for all $x \in[0,1]$. It is clear that $B_1(f) \subseteq C$. The reverse inclusion, however, is not correct. For example, let
$$g(x)= \begin{cases}1, & \text { if } x=0 \ x, & \text { if } 0<x \leq 1 .\end{cases}$$
Then $g$ belongs to $C$; but it does not belong to $B_1(f)$ since
$$d_u(f, g)=\sup {|f(x)-g(x)|: 0 \leq x \leq 1}=1 .$$

A function $f: M_1 \rightarrow M_2$ between metric spaces is continuous if and only if $f^{\leftarrow}(U)$ is an open subset of $M_1$ whenever $U$ is open in $M_2$.

Suppose $f$ is continuous. Let $U \subseteq \triangle_2$. To show that $f^{\leftarrow}(U)$ is an open subset of $M_1$, it suffices to prove that each point of $f^{\leftarrow}(U)$ is an interior point of $f^{\leftarrow}(U)$. If $a \in f^{\leftarrow}(U)$, then $f(a) \in U$. Since $f$ is continuous at $a$, the set $U$, which is a neighborhood of $f(a)$, must contain the image under $f$ of a neighborhood $V$ of $a$. But then
$$a \in V \subseteq f^{\leftarrow}\left(f^{\rightarrow}(V)\right) \subseteq f^{\leftarrow}(U)$$
which shows that $a$ lies in the interior of $f^{\leftarrow}(U)$.
Conversely, suppose that $f^{\leftarrow}(U) \subseteq M_1$ whenever $U \subseteq M_2$. To see that $f$ is continuous at an arbitrary point $a$ in $M_1$, notice that if $V$ is a neighborhood of $f(a)$, then $a \in f^{\leftarrow}(V) \subseteq M_1$. Thus $f^{\leftarrow}(V)$ is a neighborhood of $a$ whose image $f^{\rightarrow}\left(f^{\leftarrow}(V)\right)$ is contained in $V$. Thus $f$ is continuous at $a$.