Suppose $f : (a, b) \to \R$ and $x \in (a, b)$. Show that $f$ is differentiable at $x$ if and only if there is a function $h : (a, b) \to \R$ and a constant $C \in \R$ so that \[\lim_{t\to x}\frac{h(t)}{t-x} = 0 = h({\color{red}x})\] and for all $t \in (a, b)$, \[f(t) = f(x) + C(t-x) + h(t).\] Moreover, show that in this case $C = f'(x)$.
(Another equivalent condition is that there is a function $E : (a, b) \to \R$ and a constant $C \in \R$ so that $E(x) = 0$, $E$ is continuous at $x$, and \[f(t) = f(x) + C(t-x) + E(t)(t-x).\] It's probably worth thinking about why this is also equivalent, but you don't need to prove it.)
Suppose $M$ is a metric space. If $x, y \in M$, a path (in $M$) from $x$ to $y$ is a continuous function $\gamma : [0, 1] \to M$ with $\gamma(0) = x$ and $\gamma(1) = y$.
Define a relation $\sim_p$ on $M$ by $x \sim_p y$ if and only if there is a path from $x$ to $y$ in $M$. As before, set \[ [x]_{\sim_p} = \set{y \in M | x \sim_p y}.\]
The equivalence classes $[x]_{\sim_p} \subseteq M$ are called path components of $M$. If $M$ has exactly one path component, it is called path connected. (Note that $\emptyset$ is not path connected: it has zero path components, not one.)
Suppose that $E \subseteq \R^n$ is open. Show that the path components of $E$ are open (in $\R^n$).
(If you find it useful, you may use without proof the fact that functions of the form \begin{align*}f : \R &\longrightarrow \R^n \\ t &\longmapsto \vec{a} + t\vec{b}\end{align*} are continuous, where $\vec{a}, \vec{b} \in \R^n$.)
It is not true that connected sets are path connected in general. For example, the set \[T = \set{(0, y) \mid -1 \leq y \leq 1} \cup \set{\paren{x, \sin\paren{\frac1x}} \mid x \gt 0} \subseteq \R^2\] can be shown to be connected but not path connected; its path components are the two separate pieces in the presentation above.
Give an example of continuous functions $f, g : \R \to \R$ so that:
Let $S, C : \R \to [-1, 1]$ be differentiable functions with the following properties:
Let \begin{align*} f : \R &\longrightarrow \R \\ t &\longmapsto \begin{cases} 0 & \text{ if } t = 0 \\ t^2S\paren{\frac1t} & \text{ if } t \neq 0.\end{cases} \end{align*}
(Although we have not finished the proofs yet, you may assume that the Chain Rule holds, and that $h : t \mapsto \frac1t$ is differentiable on $\R\setminus\set0$ with derivative $h'(t) = \frac{-1}{t^2}$.)
Suppose $(a_n)_n$ is a sequence in $\R_\infty$. We define the limit superior of $(a_n)_n$ to be the quantity \[\limsup_{n\to\infty} a_n = \inf \set{ \sup\set{ a_k \mid k \gt n } \mid n \in \N } \in \R_\infty.\]
(Due to an error in its statement, this problem will not be graded.)
Suppose $t \in \R_\infty$. Show that $t \gt \limsup_{n\to \infty} a_n$ if and only if there are finitely many $n \in \N$ for which $t \lt a_n$.(Due to an error in its statement, this problem will not be graded.)
Suppose $t \in \R_\infty$. Show that $t \lt \limsup_{n\to \infty} a_n$ if and only if there are infinitely many $n \in \N$ for which $t \lt a_n$.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Let \begin{align*}f : \R &\longrightarrow \R \\ x &\longmapsto \begin{cases}0 & \text{ if } x \in \Q \\ x^2 & \text{ otherwise}.\end{cases}\end{align*} Show that $f$ is differentiable at $0$, and find its derivative there. (Notice that $f$ is discontinuous everywhere except at $0$! (That is an exclaimation point, not a factorial; $f$ is discontinuous at $1$.))
Suppose that $f, g : \R \to \R$ are such that $f(0) = g(0)$ and $f'(x) \lt g'(x)$ for all $x \in \R$. Prove that $\abs{f(x)} \leq \abs{g(x)}$ for all $x \in \R$, with equality only when $x = 0$.
Suppose $f : \R \to \R$. Define a function \begin{align*} F : \R^2 \setminus \set{(x, x) \mid x \in \R} &\longrightarrow \R \\ (x, y) &\longmapsto \frac{f(x) - f(y)}{x-y}. \end{align*} Under what conditions does $F$ extend continuously to $\R^2$, in the sense of Assignment 9 Question 1?
(While this certainly seems to have something to do with differentiability, it is a little subtle. Notice, for example, that taking a derivative corresponds to holding one variable fixed and letting the other tend to it, or approaching the missing diagonal horizontally or vertically only; meanwhile continuity requires everything in an open ball to be close to the correct value, not just along the two axes. (It just now occurs to me how "axis" and "axe" both have the plural "axes".))
Let us say that a function $f : \R \to \R$ has the "intermediate value property" if and only if for any $a \lt b$ in $\R$, and any $y$ between $f(a)$ and $f(b)$, there is $c \in (a, b)$ so that $f(c) = y$. The Intermediate Value Theorem tells us that continuous functions have the intermediate value property.