Advanced Math Math Analysis |02 Limits

Advanced Math & Math Analysis |02 Limits ⭐

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Let {xk}\{x_k\}{xk​} be a sequence of points in ${\mathbb{R}}^n$. We say that {xk}\{x_k\}{xk​} converges to $x$ if for every $\epsilon >0$, there exists a positive integer $N\ge 1$ such that for all $k\ge N$, the inequality $|x_k-x|<\epsilon$ holds. Denoted as
$$\underset{k\to +\infty }{\lim }{x}_k=x$$

Let ${\mathbf{x}}_0$ be an accumulation point of a set $E\subseteq {\mathbb{R}}^n$, and let $A\in {\mathbb{R}}^m$ be a given vector. A vector-valued function $f:E\to {\mathbb{R}}^m$ is said to have a limit $A$ as $\mathbf{x}$ approaches ${\mathbf{x}}_0$ if for every $\epsilon >0$, there exists a $\delta >0$ such that whenever $0<|\mathbf{x}-{\mathbf{x}}_0|<\delta$ and $\mathbf{x}\in E$, the inequality $|f(\mathbf{x})-A|<\epsilon$ holds. When the set $E$ is clear from the context, we can simply write ${\lim }_{\mathbf{x}\to {\mathbf{x}}_0}f(\mathbf{x})=A$.Denoted as
$$\underset{\begin{array}{c}\mathbf{x}\to {\mathbf{x}}_0\\ \mathbf{x}\in E\end{array}}{\lim }f(\mathbf{x})=A$$

ExampleLet $x_1>0$,$x_{n+1}=1+\frac{1}{x_n}$,$n\in {\mathbb{N}}^{\ast }$,Prove :
$$\underset{n\to \infty }{\lim }{x}_n<\infty$$

Prove

Easy to prove x_n is bounded.

$$\begin{array}{rl}{x}_{n+4}-x_{n+2}& =-\frac{1}{x_{n+1}{x}_{n+3}}(x_{n+3}-x_{n+1})\\ & =\frac{1}{x_n{x}_{n+1}{x}_{n+2}{x}_{n+3}}(x_{n+2}-x_n),\forall n\ge 1\end{array}$$

So $x_{2n}$ and $x_{2n-1}$ is monotonic and bounded sequence.So the limits is existence:

$$\begin{array}{rl}& \{\begin{array}{l}{\lim }_{n\to \infty }{x}_{2n}={\lim }_{n\to \infty }1+\frac{1}{x_{2n-1}}\\ {\lim }_{n\to \infty }{x}_{2n+1}={\lim }_{n\to \infty }1+\frac{1}{x_{2n}}\end{array}\\ \Rightarrow & \{\begin{array}{l}A=1+\frac{1}{B}\\ B=1+\frac{1}{A}\end{array}\Rightarrow A=B\end{array}$$

So the limits exists:

$$\underset{n\to \infty }{\lim }{x}_n=\frac{\sqrt{5}+1}{2}$$

ExampleGiven that the sequence {an}\{a_n\}{an​} satisfies the condition $\underset{n\to +\infty }{\lim }(a_n-a_{n-2})=0$, prove that:
$$\underset{n\to +\infty }{\lim }\frac{a_n-a_{n-1}}{n}=0$$

Prove: Let
$$b_n=\frac{a_n-a_{n-1}}{n}$$

Then, by using the Stolz Theorem, we can obtain:
$$\begin{array}{rl}\underset{n\to +\infty }{\lim }{b}_{2n}& =\underset{n\to +\infty }{\lim }\frac{a_{2n}-a_{2n-1}}{2n}\\ & =\underset{n\to +\infty }{\lim }\frac{(a_{2n}-a_{2n-1})-(a_{2n-2}-a_{2n-3})}{2n-2(n-1)}\\ & =\frac{1}{2}\underset{n\to +\infty }{\lim }\left((a_{2n}-a_{2n-2})-(a_{2n-1}-a_{2n-3})\right)=0\end{array}$$

Similarly, we can obtain $\underset{n\to +\infty }{\lim }{b}_{2n-1}=0$. Therefore,
$$\underset{n\to +\infty }{\lim }{b}_n=\underset{n\to +\infty }{\lim }\frac{a_n-a_{n-1}}{n}=0$$

Example Given ${\lim }_{nto\infty }{a}_n=A$,$b_n>0$.Donted:
$$c_n=\frac{{\sum }_{i=1}^n{a}_i{b}_i}{{\sum }_{i=1}^n{b}_i}$$
Please prove $ \lim\limits_{n\to \infty}c_n< \infty $.

Prove

If $\underset{n\to \infty }{\lim }{b}_n=B$,we have

$$\underset{n\to \infty }{\lim }{c}_n=\underset{n\to \infty }{\lim }\frac{{\sum }_{i=1}^n{a}_i{b}_i}{{\sum }_{i=1}^n{b}_i}=\underset{n\to \infty }{\lim }\frac{a_n{b}_n}{b_n}=\underset{n\to \infty }{\lim }{a}_n=A$$

If $\underset{n\to \infty }{\lim }{b}_n=\infty$,we have

$$\underset{n\to \infty }{\lim }{c}_n=\underset{n\to \infty }{\lim }\frac{{\sum }_{i=1}^n{a}_i{b}_i}{{\sum }_{i=1}^n{b}_i}=\underset{n\to \infty }{\lim }\frac{\frac{{\sum }_{i=1}^n{a}_i{b}_i}{n}}{\frac{{\sum }_{i=1}^n{b}_i}{n}}=\frac{AB}{B}=A$$

ExampleLet $n$ and $v$ be positive integers, $1\le v\le n$. Denote $n_v$ as the remainder when $n$ is divided by $v$. Calculate

(i) $$\underset{n\to +\infty }{\lim }\frac{1}{n}\left(\frac{n_1}{1}+\frac{n_2}{2}+\frac{n_3}{3}+\cdots +\frac{n_n}{n}\right)$$

(ii) $$\underset{n\to +\infty }{\lim }\frac{n_1+n_2+\cdots +n_n}{n^2}$$.

Solution: Notice that

$$n=v\left[\frac{n}{v}\right]+n_v$$

(i)We have

$$\begin{array}{rl}\underset{n\to \infty }{\lim }\frac{{\sum }_{i=1}^n\frac{n_i}{i}}{n}& =\underset{n\to \infty }{\lim }\frac{{\sum }_{i=1}^n\frac{n_i}{i}}{n}\\ & ={\int }_0^1\left(\frac{1}{x}-\left[\frac{1}{x}\right]\right)\mathrm{d}x\\ & =\sum _{n=1}^{\infty }{\int }_{\frac{1}{n+1}}^{\frac{1}{n}}\left(\frac{1}{x}-\left[\frac{1}{x}\right]\right)\mathrm{d}x\\ & =\sum _{n=1}^{\infty }\left(\ln \left(\frac{n+1}{n}\right)-\frac{1}{n+1}\right)\\ & =\underset{n\to \infty }{\lim }\left(\ln n-\sum _{k=1}^{n-1}\frac{1}{k+1}\right)=1-\gamma \end{array}$$

(ii)

We have

$$\begin{array}{rl}& \underset{n\to +\infty }{\lim }\frac{n_1+n_2+\cdots +n_n}{n^2}=\underset{n\to +\infty }{\lim }\frac{1}{n}\sum _{v=1}^n\left(1-\frac{v}{n}\lfloor \frac{n}{v}\rfloor \right)=1-{\int }_0^1\lfloor \frac{1}{x}\rfloor x\mathrm{d}x\\ & =1-\sum _{n=1}^{\infty }\frac{n}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1{)}^2}\right)=1-\frac{1}{2}\sum _{n=1}^{\infty }\left(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{(n+1{)}^2}\right)\\ & =1-\frac{1}{2}\left(1+\sum _{n=1}^{\infty }\frac{1}{(n+1{)}^2}\right)=1-\frac{{\pi }^2}{12}.\end{array}$$

Example$\alpha ,\beta ,\delta$ is positive number,Caulate:

$$\underset{n\to \infty }{\lim }\prod _{k=0}^{n-1}\frac{1+\frac{\alpha +k\delta }{n}}{1+\frac{\beta +k\delta }{n}}=(1+\delta {)}^{\frac{\alpha -\beta }{\delta }}$$

Prove

$$
\begin{align*}
\lim_{n\to\infty}\prod_{k=0}^{n-1}\frac{1+\frac{\alpha+k\delta}{n}}{1+\frac{\beta+k\delta}{n}}&=
\exp\left{\lim_{n\to\infty}\sum_{k=0}^{n-1}\ln\frac{1+\frac{\alpha+k\delta}{n}}{1+\frac{\beta+k\delta}{n}}\right}\
&=\exp\left{\lim_{n\to\infty}\sum_{k=0}^{n-1}\ln\left(1+\frac{\frac{\alpha-\beta}{\delta}}{1+\frac{\beta +k\delta}{n}}\frac{\delta}{n}\right)\right}
\
&=\exp\left{\int_{0}^{\delta}\frac{\frac{\alpha-\beta}{\delta}}{1+x}\mathrm{d}x\right}

\
&=(1+\delta)^{\frac{\alpha-\beta}{\delta}}
\end{align*}
$$

Example Caculate
$$\underset{n\to \infty }{\lim }{\int }_0^1\frac{1}{1+{\left(1+\frac{x}{n}\right)}^n}\mathrm{d}x$$

Prove

Easily to prove :

$$\frac{1}{1+{\left(1+\frac{x}{n}\right)}^n}\rightrightarrows \frac{1}{1+e^x}$$

So:

$$\underset{n\to \infty }{\lim }{\int }_0^1\frac{1}{1+{\left(1+\frac{x}{n}\right)}^n}\mathrm{d}x={\int }_0^1\frac{1}{1+e^x}\mathrm{d}x$$

Example Let $\alpha >0$ and $n{x}_n=1+o(n^{-\alpha })(n\to +\infty )$. Prove that the sequence {x1+x2+⋯+xn−ln⁡n}\{x_1 + x_2 + \cdots + x_n-\ln n\}{x1​+x2​+⋯+xn​−lnn} converges.

Proof
Denote $y_n=x_1+x_2+\cdots +x_n-\ln n(n\ge 1)$.
For any $n\ge 1$, by the Mean Value Theorem, there exists $\theta \in (0,1)$ such that
$$\begin{array}{rl}|y_{n+1}-y_n|& =|x_{n+1}-\ln (n+1)+\ln n|\\ & =\left|\frac{1}{n+1}+o((n+1{)}^{-1-\alpha })-\frac{1}{n+\theta }\right|\\ & =\left|\frac{\theta -1}{(n+1)(n+\theta )}+o((n+1{)}^{-1-\alpha })\right|\\ & \le \frac{2}{n^2}+|o((n+1{)}^{-1-\alpha })|\end{array}$$

Then, the series ${\sum }_{n=1}^{\infty }(y_{n+1}-y_n)$ converges. The sum of the first $n$ terms of this series is $y_{n+1}-y_1$, so the sequence {yn}\{y_n\}{yn​} converges.

Example$a_i,b_i$ is real nnumber,calculate:

$$\underset{x\to \infty }{\lim }\frac{1-{\prod }_{i=1}^n{\cos }^{b_i}{a}_{i}x}{x^2}$$

Solve
$$\begin{array}{rl}\underset{x\to \infty }{\lim }\frac{1-{\prod }_{i=1}^n{\cos }^{b_i}{a}_{i}x}{x^2}& =\underset{x\to \infty }{\lim }\frac{1-\exp {\sum }_{i=1}^n{b}_i\ln \cos a_{i}x}{x^2}\\ & =\underset{x\to \infty }{\lim }\frac{1-\exp {\sum }_{i=1}^n{b}_i\ln \cos a_{i}x}{x^2}\\ & =\underset{x\to \infty }{\lim }\frac{-{\sum }_{i=1}^n{b}_i\ln \cos a_{i}x}{x^2}\\ & \stackrel{L^{\prime }Hospital}{=}\sum _{i=1}^n\frac{a_i^2{b}_i}{2}\end{array}$$

ExampleLet $f^{\prime }$ be uniformly continuous on $[0,+\infty )$ and ${\lim }_{x\to +\infty }f(x)$ exists. Prove that ${\lim }_{x\to +\infty }{f}^{\prime }(x)=0$.

Prove

First, assume that ${\lim }_{x\to +\infty }{f}^{\prime }(x)\ne 0$. Then there exists ${ϵ}_0>0$ and a sequence of numbers {xn}\{x_n\}{xn​} with $x_n\to +\infty$ as $n\to \infty$, such that $f^{\prime }(x_n)\ge {ϵ}_0$ or $f^{\prime }(x_n)\le -{ϵ}_0$. Without loss of generality, we can just consider the case where $f^{\prime }(x_n)\ge {ϵ}_0$. Since $f^{\prime }(x)$ is uniformly continuous, there exists $\delta >0$ such that for all $x,x^{\prime }\ge 0$ with $|x-x^{\prime }|\le \delta$, we have $|f^{\prime }(x)-f^{\prime }(x^{\prime })|\le \frac{{ϵ}_0}{2}$. Then we can get ${ϵ}_0-f^{\prime }(y)\le f^{\prime }(x_n)-f^{\prime }(y)\le \frac{{ϵ}_0}{2}$, that is, $f^{\prime }(y)\ge \frac{{ϵ}_0}{2}$. Now we have:
$$f(x_n)-f(x_n-\delta )={\int }_{x_n-\delta }^{x_n}{f}^{\prime }(y)dy\ge {\int }_{x_n-\delta }^{x_n}\frac{{ϵ}_0}{2}dy=\frac{{ϵ}_0\delta }{2}>0$$

But as $x_n\to +\infty$, we have $0\ge \frac{{ϵ}_0\delta }{2}$, which is a contradiction. Therefore, it is proved that ${\lim }_{x\to +\infty }{f}^{\prime }(x)=0$.