 Karl Weierstrass‘s father, Wilhelm Weierstrass, was secretary to the mayor of Ostenfelde at the time of Karl’s birth. Wilhelm Weierstrass was a well educated man who had a broad knowledge of the arts and of the sciences. He certainly was well capable of attaining higher positions than he did, and this attitude may have been one of the reasons that Karl Weierstrass’s early career was in posts well below his outstanding ability. Weierstrass’s mother was Theodora Vonderforst and Karl was the eldest of Theodora and Wilhelm’s four children, none of whom married.Wilhelm Weierstrass became a tax inspector when Karl was eight years old. This job involved him in only spending short periods in any one place so Karl frequently moved from school to school as the family moved around Prussia. In 1827 Karl’s mother Theodora died and one year later his father Wilhelm remarried. By 1829 Wilhelm Weierstrass had become an assistant at the main tax office in Paderborn, and Karl entered the Catholic Gymnasium there. Weierstrass excelled at the Gymnasium despite having to take on a part-time job as a bookkeeper to help out the family finances.

# Understanding these basic definition and statement~

• Let $f: D \rightarrow \mathbb{R}$ and let $c$ be an accumulation point of $D .$ We say that a real number $L$ is a limit of $f$ at $c$, and write
$$\lim _{x \rightarrow c} f(x)=L,$$
if for each $\varepsilon>0$ there exists a $\delta>0$ such that $|f(x)-L|<\varepsilon$ for all points $x \in D$ for which $0<|x-c|<\delta$
• Monotone function: A function $f: A \rightarrow \mathbb{R}$ is increasing on $A$ if $f(x) \leq f(y)$ whenever $x<y$ and decreasing if $f(x) \geq f(y)$ whenever $x<y$ in $A$. A monotone function is one that is either increasing or decreasing.
• Bounded function: Let $f: A \rightarrow \mathbb{R}$ be a function and $B \subseteq A$. We say $f$ is bounded on $B$ if $f(B)$ is bounded, where $f(B)$ is the range of $f$ over $B$, i.e., $f(B)={f(x): x \in B}$.
• One-sided limits: Suppose the domain of $f$ is an interval $(a, b)$, then the right-hand limit of $f(x)$ at $a$ written by $\lim {x \rightarrow a^{+}} f(x)$ and $\lim {x \rightarrow a^{+}} f(x)=L$ if and only if for every $\varepsilon>0$ there exists a
$\delta>0$ such that $|f(x)-L|<\varepsilon$ whenever $x \in(a, b)$ and $a<x<a+\delta$. Similarly, the left-hand
limit of $f$ at $b$ is given by $\lim {x \rightarrow b^{-}} f(x)=L$ if and only if for every $\varepsilon>0$ there exists a $\delta>0$ such that $|f(x)-L|<\varepsilon$ whenever $x \in(a, b)$ and $b-\delta{x \rightarrow c} f(x)=L$ if and only if both one-sided limits exist and are equal to $L$.
• If $f$ is an increasing function on the interval $(a, b)$, then one-sided limits of $f$ exist at each point $c \in(a, b)$, and
$$\lim {x \rightarrow c^{-}} f(x)=L \leq f(c) \leq \lim {x \rightarrow c^{+}} f(x)=M$$
For decreasing functions, the above inequalities are reversed.
• Let $f:(b, \infty) \rightarrow \mathbb{R}$, where $b \in \mathbb{R}$. We say that $L \in \mathbb{R}$ is the limit of $f$ as $x \rightarrow \infty$, and we write
$$\lim _{x \rightarrow \infty} f(x)=L$$
provided that for each $\varepsilon>0$ there exists a real number $N>b$ such that $x>N$ implies that $|f(x)-L|<\varepsilon$
• Let $f:(b, \infty) \rightarrow \mathbb{R}$. We say that $f$ tends to $\infty$ as $x \rightarrow \infty$, and we write
$$\lim _{x \rightarrow \infty} f(x)=\infty,$$
provided that given any $M \in \mathbb{R}$ there exists a $N>b$ such that $x>N$ implies that $f(x)>M$.

# Try the following problem to test yourself！

Problem 1.

Let $\left{x_{n}\right}$ and $\left{y_{n}\right}$ be two real sequences such that
(a) $x_{n} \leq y_{n}$ for all $n$;
(

Evaluate the limit
$$\lim _{x \rightarrow 0} \frac{\sin (x)}{\sqrt{1-\cos (x)}}$$

Proof .

Since $\left{y_{n}\right}$ is decreasing, we have $y_{n} \leq y_{1}$ for $n \geq 1 .$ So for any $n \geq 1$ we have $x_{n} \ The square root in the denominator is the part complicating this limit. In order to get rid of this problem, we will use the following trigonometric identities: $$\cos (\theta)=1-2 \sin ^{2}\left(\frac{\theta}{2}\right) \quad \text { and } \quad \sin (\theta)=2 \cos \left(\frac{\theta}{2}\right) \sin \left(\frac{\theta}{2}\right) \text {. }$$ So $$\frac{\sin (x)}{\sqrt{1-\cos (x)}}=\frac{2 \cos \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)}{\sqrt{2 \sin ^{2}\left(\frac{x}{2}\right)}}=\frac{2 \cos \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right)}{\sqrt{2}\left|\sin \left(\frac{x}{2}\right)\right|} .$$ First note that $$\lim {x \rightarrow 0} \frac{2 \cos \left(\frac{x}{2}\right)}{\sqrt{2}}=\sqrt{2} \text {. }$$ So let us focus on the limit$\lim {x \rightarrow 0} \frac{\sin \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right) \mid}$. In order to get rid of the absolute value, we will consider the limits to the right of 0 and to the left of 0 . From the properties of the sin function we know that for any$x \in(0, \pi)$, then$\sin \left(\frac{x}{2}\right)>0$and if$x \in(-\pi, 0)$, then$\sin \left(\frac{x}{2}\right)<0$. Hence $$\lim {x \rightarrow 0+} \frac{\sin \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right) \mid}=\lim {x \rightarrow 0+} \frac{\sin \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}=1$$ and $$\lim {x \rightarrow 0-} \frac{\sin \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right) \mid}=\lim {x \rightarrow 0+} \frac{\sin \left(\frac{x}{2}\right)}{-\sin \left(\frac{x}{2}\right)}=-1$$ Putting all the previous information together we get $$\lim {x \rightarrow 0+} \frac{\sin (x)}{\sqrt{1-\cos (x)}}=\sqrt{2} \text { and } \lim {x \rightarrow 0-} \frac{\sin (x)}{\sqrt{1-\cos (x)}}=-\sqrt{2} \text {. }$$ This obviously implies that the limit$\lim _{x \rightarrow 0} \frac{\sin (x)}{\sqrt{1-\cos (x)}}$does not exist. Problem 2. Let$f, g: D \rightarrow \mathbb{R}$. Let$x_{0} \in D$such that$\lim {x \rightarrow x{0}} f(x)$and$\lim {x \rightarrow x{0}} g(x)$exist. Discuss the existence of the limits $$\lim {x \rightarrow x{0}} \min {f(x), g(x)}, \text { and } \lim {x \rightarrow x{0}} \max {f(x), g(x)} .$$ Proof . We will make use of the identities $$\max {a, b}=\frac{a+b+|a-b|}{2} \text { and } \min {a, b}=\frac{a+b-|a-b|}{2}$$ for any$a, b \in \mathbb{R}$. Now set $$\lim {x \rightarrow x{0}} f(x)=l \text { and } \lim {x \rightarrow x{0}} g(x)=L$$ Properties on the limits of functions will imply$\lim {x \rightarrow x{0}} f(x)-g(x)=l-L .$In Problem$5.1$, we showed that$\lim {x \rightarrow x{0}}|f(x)-g(x)|=|l-L| .$Hence $$\lim {x \rightarrow x{0}} \frac{f(x)+g(x)+|f(x)-g(x)|}{2}=\frac{l+L+|l-L|}{2},$$ or $$\lim {x \rightarrow x{0}} \max {f(x), g(x)}=\max {l, L}$$ A similar proof will give $$\lim {x \rightarrow x{0}} \min {f(x), g(x)}=\min {l, L}$$ It is obvious that one may generalize these conclusions to a finite number of functions. But the infinite case is not true. Indeed, take $$f_{n}(x)= \begin{cases}x^{n} & \text { if } x \in[0,1], \ 1 & \text { if } x \geq 1\end{cases}$$ Then it is quite easy to check that$\lim {x \rightarrow 1} f{n}(x)=1 .$But if we set$f(x)=\inf {n \geq 1} f{n}(x)$, then we have $$f(x)= \begin{cases}0 & \text { if } x \in[0,1], \ 1 & \text { if } x \geq 1\end{cases}$$ Hence $$\lim {x \rightarrow 1-} f(x)=0 \text { and } \lim {x \rightarrow 1+} f(x)=1$$ Problem 3. Let$f: \mathbb{R} \rightarrow \mathbb{R}$such that $$f(x+y)=f(x)+f(y)$$ for any$x, y \in \mathbb{R}$. Assume that$\lim _{x \rightarrow 0} f(x)=f(0)$. Find$f(x)$. Proof . The equation satisfied by$f(x)$is usually known as a functional equation. First note that$f(n x)=n f(x)$for any$n \in \mathbb{N}$. Indeed, we have$f(1 \cdot x)=1 \cdot f(x)$. Assume that$f(n x)=n f(x)$. Then $$f((n+1) x)=f(n x+x)=f(n x)+f(x)=n f(x)+f(x)=(n+1) f(x) .$$ By induction we get the desired identity. Let$r=\frac{p}{q} \in \mathbb{Q}$. Then we have$f(q r x)=f(p x)=p f(x)$, and since$f(q r x)=q f(r x)$, we get$f(r x)=\frac{p}{q} f(x)=r f(x) .$In particular, we have$f(r)=r f(1)$. Let$x \in \mathbb{R}$. Then there exists a sequence of rational numbers$\left{r_{n}\right}$such that$\lim r_{n}=x$. Since$f(x)=f\left(x-r_{n}+r_{n}\right)=f\left(x-r_{n}\right)+f\left(r_{n}\right)=f\left(x-r_{n}\right)+r_{n} f(1)$, we have$f(x)=f\left(x-r_{n}\right)+r_{n} f(1)$. Now if we take the limit as$n \rightarrow \infty$, by the existence of$\lim _{x \rightarrow 0} f(x)=f(0)$, we have that$f(x)=f(0)+x f(1)$. Note that$f(x)=f(x+0)=f(x)+f(0)$for any$x \in \mathbb{R}$, so$f(0)=0\$. Therefore, we have
$$f(x)=x f(1)=m x .$$