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！
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
$$
\lim _{x \rightarrow 0} \frac{\sin (x)}{\sqrt{1-\cos (x)}}
$$
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.
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
$$
\lim {x \rightarrow x{0}} \min {f(x), g(x)}, \text { and } \lim {x \rightarrow x{0}} \max {f(x), g(x)} .
$$
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
$$
\lim {x \rightarrow 1-} f(x)=0 \text { and } \lim {x \rightarrow 1+} f(x)=1
$$
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 $f(x)$.
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
$$
f(x)=x f(1)=m x .
$$
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Fourier analysis代写
离散数学代写
Partial Differential Equations代写可以参考一份偏微分方程midterm答案解析
