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 Rn\mathbb{R}^nRn. We say that {xk}\{x_k\}{xk} converges to xxx if for every ε>0\varepsilon > 0ε>0, there exists a positive integer N≥1N \geq 1N≥1 such that for all k≥Nk \geq Nk≥N, the inequality ∣xk−x∣<ε|x_k - x| < \varepsilon∣xk−x∣<ε holds. Denoted as
lim⁡k→+∞xk=x\lim_{k \to +\infty} x_k = xk→+∞limxk=x
Let x0\boldsymbol{x}_0x0 be an accumulation point of a set E⊆RnE \subseteq \mathbb{R}^nE⊆Rn, and let A∈RmA \in \mathbb{R}^mA∈Rm be a given vector. A vector-valued function f:E→Rmf: E \to \mathbb{R}^mf:E→Rm is said to have a limit AAA as x\boldsymbol{x}x approaches x0\boldsymbol{x}_0x0 if for every ε>0\varepsilon > 0ε>0, there exists a δ>0\delta > 0δ>0 such that whenever 0<∣x−x0∣<δ0 < |\boldsymbol{x} - \boldsymbol{x}0| < \delta0<∣x−x0∣<δ and x∈E\boldsymbol{x} \in Ex∈E, the inequality ∣f(x)−A∣<ε|f(\boldsymbol{x}) - A| < \varepsilon∣f(x)−A∣<ε holds. When the set EEE is clear from the context, we can simply write lim⁡x→x0f(x)=A\lim{\boldsymbol{x} \to \boldsymbol{x}0} f(\boldsymbol{x}) = Alimx→x0f(x)=A.Denoted as
lim⁡x→x0x∈Ef(x)=A\lim{\substack{\boldsymbol{x} \to \boldsymbol{x}_0 \\ \boldsymbol{x} \in E}} f(\boldsymbol{x}) = Ax→x0x∈Elimf(x)=A

Example Let x1>0x_1>0x1>0,xn+1=1+1xnx_{n+1}=1+\frac{1}{x_n}xn+1=1+xn1,n∈N∗n\in \mathbb{N}^*n∈N∗,Prove :
lim⁡n→∞xn<∞\lim_{n\to \infty}x_n< \inftyn→∞limxn<∞

Prove

Easy to prove x_n is bounded.

xn+4−xn+2=−1xn+1xn+3(xn+3−xn+1)=1xnxn+1xn+2xn+3(xn+2−xn),∀n≥1\begin{align*} 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}),\quad \forall n \ge 1 \end{align*} xn+4−xn+2=−xn+1xn+31(xn+3−xn+1)=xnxn+1xn+2xn+31(xn+2−xn),∀n≥1

So x2nx_{2n}x2n and x2n−1x_{2n-1}x2n−1 is monotonic and bounded sequence.So the limits is existence:

{lim⁡n→∞x2n=lim⁡n→∞1+1x2n−1lim⁡n→∞x2n+1=lim⁡n→∞1+1x2n⇒{A=1+1BB=1+1A⇒A=B\begin{align*} &\begin{cases} \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{cases}\\ \Rightarrow & \begin{cases} A=1+\frac{1}{B}\\ B=1+\frac{1}{A} \end{cases}\Rightarrow A=B \end{align*} ⇒{limn→∞x2n=limn→∞1+x2n−11limn→∞x2n+1=limn→∞1+x2n1{A=1+B1B=1+A1⇒A=B

So the limits exists:

lim⁡n→∞xn=5+12\lim_{n\to \infty}x_n =\frac{\sqrt{5}+1}{2}n→∞limxn=25 +1

Example Given that the sequence {an}\{a_n\}{an} satisfies the condition lim⁡n→+∞(an−an−2)=0\lim\limits_{n \to +\infty} (a_n - a_{n - 2}) = 0n→+∞lim(an−an−2)=0, prove that:
lim⁡n→+∞an−an−1n=0\lim_{n \to +\infty} \frac{a_n - a_{n - 1}}{n} = 0 n→+∞limnan−an−1=0

Prove : Let
bn=an−an−1nb_n = \frac{a_n - a_{n - 1}}{n}bn=nan−an−1

Then, by using the Stolz Theorem, we can obtain:
lim⁡n→+∞b2n=lim⁡n→+∞a2n−a2n−12n=lim⁡n→+∞(a2n−a2n−1)−(a2n−2−a2n−3)2n−2(n−1)=12lim⁡n→+∞((a2n−a2n−2)−(a2n−1−a2n−3))=0 \begin{align*} \lim_{n \to +\infty} b_{2n} &= \lim_{n \to +\infty} \frac{a_{2n} - a_{2n - 1}}{2n}\\ &= \lim_{n \to +\infty} \frac{(a_{2n} - a_{2n - 1}) - (a_{2n - 2} - a_{2n - 3})}{2n - 2(n - 1)}\\ &= \frac{1}{2} \lim_{n \to +\infty} \left((a_{2n} - a_{2n - 2}) - (a_{2n - 1} - a_{2n - 3})\right) = 0 \end{align*} n→+∞limb2n=n→+∞lim2na2n−a2n−1=n→+∞lim2n−2(n−1)(a2n−a2n−1)−(a2n−2−a2n−3)=21n→+∞lim((a2n−a2n−2)−(a2n−1−a2n−3))=0

Similarly, we can obtain lim⁡n→+∞b2n−1=0\lim\limits_{n \to +\infty} b_{2n - 1} = 0n→+∞limb2n−1=0. Therefore,
lim⁡n→+∞bn=lim⁡n→+∞an−an−1n=0\lim\limits_{n \to +\infty} b_n = \lim\limits_{n \to +\infty} \frac{a_n - a_{n - 1}}{n} = 0n→+∞limbn=n→+∞limnan−an−1=0

Example Given lim⁡n to∞an=A\lim_{n\ to\infty }a_n=Alimn to∞an=A,bn>0b_n>0bn>0.Donted:
cn=∑i=1naibi∑i=1nbic_n=\frac{\sum_{i=1}^na_ib_i}{\sum_{i=1}^nb_i}cn=∑i=1nbi∑i=1naibi

Please prove \\lim\\limits_{n\\to \\infty}c_n\< \\infty .

Prove

If lim⁡n→∞bn=B\lim\limits_{n\to \infty} b_n=Bn→∞limbn=B,we have

lim⁡n→∞cn=lim⁡n→∞∑i=1naibi∑i=1nbi=lim⁡n→∞anbnbn=lim⁡n→∞an=A\lim\limits_{n\to \infty}c_n=\lim\limits_{n\to \infty}\frac{\sum_{i=1}^na_ib_i}{\sum_{i=1}^nb_i}=\lim\limits_{n\to \infty}\frac{a_nb_n}{b_n}=\lim\limits_{n\to \infty}a_n=An→∞limcn=n→∞lim∑i=1nbi∑i=1naibi=n→∞limbnanbn=n→∞liman=A

If lim⁡n→∞bn=∞\lim\limits_{n\to \infty} b_n=\inftyn→∞limbn=∞,we have

lim⁡n→∞cn=lim⁡n→∞∑i=1naibi∑i=1nbi=lim⁡n→∞∑i=1naibin∑i=1nbin=ABB=A\lim\limits_{n\to \infty}c_n=\lim\limits_{n\to \infty}\frac{\sum_{i=1}^na_ib_i}{\sum_{i=1}^nb_i}=\lim\limits_{n\to \infty} \frac{\frac{\sum_{i=1}^na_ib_i}{n}}{\frac{\sum_{i=1}^nb_i}{n}}=\frac{AB}{B}=An→∞limcn=n→∞lim∑i=1nbi∑i=1naibi=n→∞limn∑i=1nbin∑i=1naibi=BAB=A

Example Let nnn and vvv be positive integers, 1≤v≤n1 \leq v \leq n1≤v≤n. Denote nvn_vnv as the remainder when nnn is divided by vvv. Calculate

(i) lim⁡n→+∞1n(n11+n22+n33+⋯+nnn)\lim_{n \to +\infty} \frac{1}{n} \left( \frac{n_1}{1} + \frac{n_2}{2} + \frac{n_3}{3} + \cdots + \frac{n_n}{n} \right)n→+∞limn1(1n1+2n2+3n3+⋯+nnn)

(ii) lim⁡n→+∞n1+n2+⋯+nnn2\lim_{n \to +\infty} \frac{n_1 + n_2 + \cdots + n_n}{n^2}n→+∞limn2n1+n2+⋯+nn.

Solution: Notice that

n=v[nv]+nvn = v \left[\frac{n}{v} \right] + n_v n=v[vn]+nv

(i)We have

lim⁡n→∞∑i=1nniin=lim⁡n→∞∑i=1nniin=∫01(1x−[1x])dx=∑n=1∞∫1n+11n(1x−[1x])dx=∑n=1∞(ln⁡(n+1n)−1n+1)=lim⁡n→∞(ln⁡n−∑k=1n−11k+1)=1−γ\begin{align*}\lim_{n\to \infty } \frac{\sum_{i=1}^n\frac{n_i}{i}}{n}&=\lim_{n\to \infty } \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)\\&=\lim_{n\to \infty}\left(\ln n - \sum_{k=1}^{n-1}\frac{1}{k+1}\right)=1-\gamma \end{align*}n→∞limn∑i=1nini=n→∞limn∑i=1nini=∫01(x1−[x1])dx=n=1∑∞∫n+11n1(x1−[x1])dx=n=1∑∞(ln(nn+1)−n+11)=n→∞lim(lnn−k=1∑n−1k+11)=1−γ

(ii)

We have

lim⁡n→+∞n1+n2+⋯+nnn2=lim⁡n→+∞1n∑v=1n(1−vn⌊nv⌋)=1−∫01⌊1x⌋xdx=1−∑n=1∞n2(1n2−1(n+1)2)=1−12∑n=1∞(1n−1n+1+1(n+1)2)=1−12(1+∑n=1∞1(n+1)2)=1−π212. \begin{align*} &\lim_{n \to +\infty} \frac{n_1 + n_2 + \cdots + n_n}{n^2} = \lim_{n \to +\infty} \frac{1}{n} \sum_{v = 1}^{n} \left( 1 - \frac{v}{n} \left\lfloor \frac{n}{v} \right\rfloor \right) = 1 - \int_{0}^{1} \left\lfloor \frac{1}{x} \right\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{align*} n→+∞limn2n1+n2+⋯+nn=n→+∞limn1v=1∑n(1−nv⌊vn⌋)=1−∫01⌊x1⌋xdx=1−n=1∑∞2n(n21−(n+1)21)=1−21n=1∑∞(n1−n+11+(n+1)21)=1−21(1+n=1∑∞(n+1)21)=1−12π2.

Example α,β,δ\alpha,\beta ,\deltaα,β,δ is positive number,Caulate:

lim⁡n→∞∏k=0n−11+α+kδn1+β+kδn=(1+δ)α−βδ\lim_{n\to\infty}\prod_{k=0}^{n-1}\frac{1+\frac{\alpha+k\delta}{n}}{1+\frac{\beta+k\delta}{n}}=(1+\delta)^{\frac{\alpha-\beta}{\delta}}n→∞limk=0∏n−11+nβ+kδ1+nα+kδ=(1+δ)δα−β

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
lim⁡n→∞∫0111+(1+xn)ndx\lim_{n\to \infty}\int_{0}^1\frac{1}{1+\left(1+\frac{x}{n}\right)^n}\mathrm{d}xn→∞lim∫011+(1+nx)n1dx

Prove

Easily to prove :

11+(1+xn)n⇉11+ex\frac{1}{1+\left(1+\frac{x}{n}\right)^n} \rightrightarrows \frac{1}{1+e^x}1+(1+nx)n1⇉1+ex1

So:

lim⁡n→∞∫0111+(1+xn)ndx=∫0111+exdx\lim_{n\to \infty}\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}xn→∞lim∫011+(1+nx)n1dx=∫011+ex1dx

Example Let α>0\alpha\gt0α>0 and nxn=1+o(n−α)(n→+∞)nx_n = 1 + o(n^{-\alpha})(n \to +\infty)nxn=1+o(n−α)(n→+∞). 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 yn=x1+x2+⋯+xn−ln⁡n(n≥1)y_n = x_1 + x_2 + \cdots + x_n-\ln n(n\geq1)yn=x1+x2+⋯+xn−lnn(n≥1).

For any n≥1n\geq1n≥1, by the Mean Value Theorem, there exists θ∈(0,1)\theta\in(0,1)θ∈(0,1) such that
∣yn+1−yn∣=∣xn+1−ln⁡(n+1)+ln⁡n∣=∣1n+1+o((n+1)−1−α)−1n+θ∣=∣θ−1(n+1)(n+θ)+o((n+1)−1−α)∣≤2n2+∣o((n+1)−1−α)∣ \begin{align*} |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|\\ &\leq\frac{2}{n^2}+|o((n + 1)^{-1-\alpha})| \end{align*} ∣yn+1−yn∣=∣xn+1−ln(n+1)+lnn∣= n+11+o((n+1)−1−α)−n+θ1 = (n+1)(n+θ)θ−1+o((n+1)−1−α) ≤n22+∣o((n+1)−1−α)∣

Then, the series ∑n=1∞(yn+1−yn)\sum_{n = 1}^{\infty}(y_{n + 1}-y_n)∑n=1∞(yn+1−yn) converges. The sum of the first nnn terms of this series is yn+1−y1y_{n + 1}-y_1yn+1−y1, so the sequence {yn}\{y_n\}{yn} converges.

Example ai,bia_i,b_iai,bi is real nnumber,calculate:

lim⁡x→∞1−∏i=1ncos⁡biaixx2\lim_{x\to \infty}\frac{1-\prod_{i=1}^n\cos^{b_i}{a_ix}}{x^2}x→∞limx21−∏i=1ncosbiaix

Solve
lim⁡x→∞1−∏i=1ncos⁡biaixx2=lim⁡x→∞1−exp⁡∑i=1nbiln⁡cos⁡aixx2=lim⁡x→∞1−exp⁡∑i=1nbiln⁡cos⁡aixx2=lim⁡x→∞−∑i=1nbiln⁡cos⁡aixx2=L′Hospital∑i=1nai2bi2 \begin{align*} \lim_{x\to \infty}\frac{1-\prod_{i=1}^n\cos^{b_i}{a_ix}}{x^2}&=\lim_{x\to \infty}\frac{1-\exp{\sum_{i=1}^n}{b_i}\ln{\cos{a_ix}}}{x^2}\\ &=\lim_{x\to \infty}\frac{1-\exp{\sum_{i=1}^n}{b_i}\ln{\cos{a_ix}}}{x^2}\\ &=\lim_{x\to \infty}\frac{-\sum_{i=1}^n{b_i}\ln{\cos{a_ix}}}{x^2} \\& \overset{L'Hospital}{=} \sum_{i=1}^n\frac{a_i^2b_i}{2} \end{align*} x→∞limx21−∏i=1ncosbiaix=x→∞limx21−exp∑i=1nbilncosaix=x→∞limx21−exp∑i=1nbilncosaix=x→∞limx2−∑i=1nbilncosaix=L′Hospitali=1∑n2ai2bi

Example Let f′f'f′ be uniformly continuous on [0,+∞)[0, +\infty)[0,+∞) and lim⁡x→+∞f(x)\lim_{x \to +\infty} f(x)limx→+∞f(x) exists. Prove that lim⁡x→+∞f′(x)=0\lim_{x \to +\infty} f'(x)=0limx→+∞f′(x)=0.

Prove

First, assume that lim⁡x→+∞f′(x)≠0\lim_{x \to +\infty} f'(x) \neq 0limx→+∞f′(x)=0. Then there exists ϵ0>0\epsilon_0 > 0ϵ0>0 and a sequence of numbers {xn}\{x_n\}{xn} with xn→+∞x_n \to +\inftyxn→+∞ as n→∞n \to \inftyn→∞, such that f′(xn)≥ϵ0f'(x_n) \geq \epsilon_0f′(xn)≥ϵ0 or f′(xn)≤−ϵ0f'(x_n) \leq -\epsilon_0f′(xn)≤−ϵ0. Without loss of generality, we can just consider the case where f′(xn)≥ϵ0f'(x_n) \geq \epsilon_0f′(xn)≥ϵ0. Since f′(x)f'(x)f′(x) is uniformly continuous, there exists δ>0\delta > 0δ>0 such that for all x,x′≥0x, x' \geq 0x,x′≥0 with ∣x−x′∣≤δ|x - x'| \leq \delta∣x−x′∣≤δ, we have ∣f′(x)−f′(x′)∣≤ϵ02|f'(x) - f'(x')| \leq \frac{\epsilon_0}{2}∣f′(x)−f′(x′)∣≤2ϵ0. Then we can get ϵ0−f′(y)≤f′(xn)−f′(y)≤ϵ02\epsilon_0 - f'(y) \leq f'(x_n) - f'(y) \leq \frac{\epsilon_0}{2}ϵ0−f′(y)≤f′(xn)−f′(y)≤2ϵ0, that is, f′(y)≥ϵ02f'(y) \geq \frac{\epsilon_0}{2}f′(y)≥2ϵ0. Now we have:
f(xn)−f(xn−δ)=∫xn−δxnf′(y)dy≥∫xn−δxnϵ02dy=ϵ0δ2>0 f(x_n) - f(x_n - \delta) = \int_{x_n - \delta}^{x_n} f'(y) dy \geq \int_{x_n - \delta}^{x_n} \frac{\epsilon_0}{2} dy = \frac{\epsilon_0 \delta}{2} > 0 f(xn)−f(xn−δ)=∫xn−δxnf′(y)dy≥∫xn−δxn2ϵ0dy=2ϵ0δ>0

But as xn→+∞x_n \to +\inftyxn→+∞, we have 0≥ϵ0δ20 \geq \frac{\epsilon_0 \delta}{2}0≥2ϵ0δ, which is a contradiction. Therefore, it is proved that lim⁡x→+∞f′(x)=0\lim_{x \to +\infty} f'(x) = 0limx→+∞f′(x)=0.