Usually (but not always) the sequences that arise in practice have a recog-nisable pattern and can be described by a formula. Prove that Given any number , the interval can contain at most two integers. Sometimes when calculating a limit, the answer varies depending on the path taken toward $$(a,b)$$. It is equivalent to say that for every neighbourhood $$V$$ of $$x$$ and every $$n_{0}\in \mathbb {N}$$, there is some $$n\geq n_{0}$$ such that $$x_{n}\in V$$. For example, if and is a term of a sequence, the distance between and, denoted by, is By using the concept of distance, the above informal definition can be made rigorous. Not every sequence has this behavior: those that do are called convergent, while those that don't are called divergent. A number is called the limit superior if infinitely many terms of the sequence are greater than - ε for any positive ε, while only a finite number of terms are greater than + ε. If $${\varepsilon _1},{\varepsilon _2}$$ are two arbitrary small positive numbers then it readily follows that $$l$$ is a limit point of a sequence $$u$$ if and only if $${u_n} \in \left( {l – \varepsilon ,l + \varepsilon } \right)$$ for infinitely many values of $$\eta$$. Whenever we simply write $$\varepsilon > 0$$ it is implied that $$\varepsilon$$ may be howsoever small positive number. A number l is said to be a limit point of a sequence u if every neighborhood N l of l is such that u n ∈ N l, for infinitely many values of n ∈ N, i.e. Example of Limit from Below. Evidently, if $$l = {u_n}$$ for infinitely many values of $$n$$ then $$l$$ is a limit point of the sequence $$u$$. Example of an Option Combination Fragment . Now pick the tolerance 0 In the above example this ﬁxed point is also the limiting value of the sequence. In sequences (unlike sets) an infinitely repeating term counts as infinitely many terms. for any $$\varepsilon > 0$$, $${u_n} \in \left( {l – \varepsilon ,l + \varepsilon } \right)$$, for finitely many values of $$n \in \mathbb{N}$$. Example 1.6. Find a pointwise convergent sequence of differentiable functions such that the limit function is continuous but not differentiable. Example: A bounded closed subset of is sequentially compact, by Heine-Borel Theorem. The Limit of a Sequence The concept of determining if sequence converges or diverges. Your email address will not be published. Example: Find the limit of the sequence x n = {((-1) n + n 2)/n 2} Show Step-by-step Solutions. Non-example: If a subset of a metric space is not closed, this subset can not be sequentially compact: just consider a sequence converging to a point outside of the subset! The number s is called the limit of the sequence. Definition. So by de nition, y 2S. Theorem: The set of limit points $$E$$ of every sequence $$u$$ is a closed set. 1 is an upper bound and the supremum. Then we may treat sequences and sets alike. Equivalently: every sequence has a converging sequence. Solution: For any $$\varepsilon > 0$$, $${u_n} = 1 \in \left( {1 – \varepsilon ,1 + \varepsilon } \right)$$ $$\forall n \in \mathbb{N}$$. Deﬁnition 3.1 The number L is the limit of the sequence {an} if (1) given ǫ > 0, an ≈ ǫ L for n ≫ 1. Example 3 Three examples: 1. The two notations for the limit of a sequence are: lim n→∞ {an} = L ; an → L as n → ∞ . Let a The higher is, the smaller is and the closer it gets to .Therefore, intuitively, the limit of the sequence should be : It is straightforward to prove that is indeed a limit … The limit of a sequence is the value the sequence approaches as the number of terms goes to infinity. Example of a Loop Sequence. Let’s take a look at a couple of sequences. $\lim_{n \to \infty} \frac{1}{n}-\sqrt{2}=0-\sqrt{2}=-\sqrt{2}$. Now pick the tolerance 0 The points 0 and 1 are both limit points of the interval (0, 1). A sequence is convergent if it is bounded and infL= supL. In this case, we say that x 0 is the limit of the sequence and write x n := x 0 . If this limit exists, then we say that the sequence $\{ a_n \}$ Converges, and if this limit doesn't exist then we say say the sequence $\{ a_n \}$ Diverges. A sequence with no limit is called divergent. Example: limx→10 x2 = 5. A positive number $$\eta$$ is said to be arbitrarily small if given any $$\varepsilon > 0$$, $$\eta$$ may be chosen such that $$0 < \eta < \varepsilon$$. $\lim_{n \to \infty} \frac{1}{n^2}+\frac{2}{n^3}-\frac{100}{n^6}=0+0-0=0$, Count limit of sequence $$\lim_{n \to \infty} n\left(\sqrt{2n^2+1}-\sqrt{2n^2-1}\right)$$, $\begin{split}&\lim_{n \to \infty} n\left(\sqrt{2n^2+1}-\sqrt{2n^2-1}\right)=\\[16pt] &=\lim_{n \to \infty} \frac{n\left(2n^2+1-(2n^2-1)\right)}{\left(\sqrt{2n^2+1}+\sqrt{2n^2-1}\right)}=\\[16pt] &=\lim_{n \to \infty} \frac{2n}{\left(\sqrt{2n^2+1}+\sqrt{2n^2-1}\right)}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\dfrac{2n}{n}}{\sqrt{\dfrac{2n^2}{n^2}+\dfrac{1}{n^2}}+\sqrt{\dfrac{2n^2}{n^2}-\dfrac{1}{n^2}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{2}{\sqrt{2+\dfrac{1}{n^2}}+\sqrt{2-\dfrac{1}{n^2}}}=\\[16pt] &=\frac{2}{2\sqrt{2}}=\frac{\sqrt{2}}{2}\end{split}$, Count limit of sequence $$\lim_{n \to \infty} n\left(\sqrt{7n^2+3}-\sqrt{7n^2-3}\right)$$, $\begin{split} &\lim_{n \to \infty} n\left(\sqrt{7n^2+3}-\sqrt{7n^2-3}\right)=\\[16pt] &=\lim_{n \to \infty} \frac{n\left(7n^2+3-(7n^2-3)\right)}{\left(\sqrt{7n^2+3}+\sqrt{7n^2-3}\right)}=\\[16pt] &=\lim_{n \to \infty} \frac{6n}{\left(\sqrt{7n^2+3}+\sqrt{7n^2-3}\right)}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\dfrac{6n}{n}}{\sqrt{\dfrac{7n^2}{n^2}+\dfrac{3}{n^2}}+\sqrt{\dfrac{7n^2}{n^2}-\dfrac{3}{n^2}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{6}{\sqrt{7+\dfrac{3}{n^2}}+\sqrt{7-\dfrac{3}{n^2}}}=\\[16pt] &=\frac{6}{2\sqrt{7}}=\frac{3}{\sqrt{7}}=\frac{3\sqrt{7}}{7} \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \frac{5n^6-3n^4+2}{5-9n^6}$$, $\begin{split} &\lim_{n \to \infty} \frac{5n^6-3n^4+2}{5-9n^6}=\\[15pt] &=\lim_{n \to \infty} \frac{5n^6-3n^4+2}{5-9n^6}\ \frac{:n^6}{:n^6}=\\[15pt] &=\lim_{n \to \infty} \dfrac{\dfrac{5n^6}{n^6}-\dfrac{3n^4}{n^6}+\dfrac{2}{n^6}}{\dfrac{5}{n^6}-\dfrac{9n^6}{n^6}}=\\[15pt] &=\lim_{n \to \infty} \dfrac{5-\dfrac{3}{n^2}+\dfrac{2}{n^6}}{\dfrac{5}{n^6}-9}=\\[15pt] &=-\frac{5}{9} \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \sqrt{n^2+4n+1}-\sqrt{n^2+2n}$$, $\begin{split} &\lim_{n \to \infty} \sqrt{n^2+4n+1}-\sqrt{n^2+2n}=\\[12pt] &=\lim_{n \to \infty} \frac{n^2+4n+1-n^2-2n}{\sqrt{n^2+4n+1}+\sqrt{n^2+2n}}=\\[16pt] &=\lim_{n \to \infty} \frac{2n+1}{\sqrt{n^2+4n+1}+\sqrt{n^2+2n}}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\dfrac{2n}{n}+\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{4n}{n^2}+\dfrac{1}{n^2}}+\sqrt{\dfrac{n^2}{n^2}+\dfrac{2n}{n^2}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{2+\dfrac{1}{n}}{\sqrt{1+\dfrac{4}{n}+\dfrac{1}{n^2}}+\sqrt{1+\dfrac{2}{n}}}=\\[16pt] &=\frac{2}{\sqrt{1}+\sqrt{1}}=\frac{2}{2}=1 \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \frac{1-2+3-4+...-2n}{\sqrt{n^2+1}}$$, $\begin{split} &\lim_{n \to \infty} \frac{1-2+3-4+...-2n}{\sqrt{n^2+1}}=\\[16pt] &=\lim_{n \to \infty} \frac{\left(1+3+...+(2n-1)\right)-(2+4+...+2n)}{\sqrt{n^2+1}}=\\[16pt] &\{\text{w liczniku mamy dwie sumy ciÄ gÃ³w arytmetycznych}\}\\[16pt] &=\lim_{n \to \infty} \frac{\dfrac{(2n-1)+1}{2}\cdot n-\dfrac{2n+2}{2}\cdot n}{\sqrt{n^2+1}}=\\[16pt] &=\lim_{n \to \infty} \frac{n^2-n^2-n}{\sqrt{n^2+1}}=\\[16pt] &=\lim_{n \to \infty} \frac{-n}{\sqrt{n^2+1}}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{1}{n^2}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{-1}{\sqrt{1+\dfrac{1}{n^2}}}=\\[16pt] &=\frac{-1}{\sqrt{1}}=-1 \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \dfrac{1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^n}}{1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^n}}$$, W liczniku mamy sumÄ ciÄgu geometrycznego: $1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}=\frac{1}{1-\frac{1}{2}}=2$ W mianowniku rÃ³wnieÅ¼ mamy sumÄ ciÄgu geometrycznego: $1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^n}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$ Zatem mamy: $\lim_{n \to \infty} \dfrac{1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^n}}{1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^n}}=\dfrac{2}{\frac{3}{2}}=\frac{4}{3}$, Count limit of sequence $$\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}}$$, $\begin{split} &\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}}=\\[16pt] &=\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}}\ \frac{:\sqrt{n}}{:\sqrt{n}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\sqrt{\dfrac{n}{n}}}{\sqrt{\dfrac{n}{n}+\sqrt{\dfrac{n}{n^2}+\sqrt{\dfrac{n}{n^4}}}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\sqrt{1}}{\sqrt{1+\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}}}}}=\\[16pt] &=\frac{1}{\sqrt{1}}=1 \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \sqrt{2}\cdot \sqrt{2}\cdot \sqrt{2}\cdot ...\cdot \sqrt[2^n]{2}$$, $\begin{split} &\lim_{n \to \infty} \sqrt{2}\cdot \sqrt{2}\cdot \sqrt{2}\cdot ...\cdot \sqrt[2^n]{2}=\\[16pt] &=\lim_{n \to \infty} 2^\tfrac{1}{2}\cdot 2^\tfrac{1}{4}\cdot ...\cdot 2^\tfrac{1}{2^n}=\\[16pt] &=\lim_{n \to \infty} 2^{\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+...+\tfrac{1}{2^n}}=\\[16pt] &= 2^{\tfrac{\tfrac{1}{2}}{1-\tfrac{1}{2}}}=\\[16pt] &= 2^{\tfrac{1}{2}\cdot \tfrac{2}{1}}=\\[16pt] &= 2^1=2 \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \left(\sqrt{n+6\sqrt{n}+1}-\sqrt{n}\right)$$, $\begin{split} &\lim_{n \to \infty} \left(\sqrt{n+6\sqrt{n}+1}-\sqrt{n}\right)=\\[16pt] &=\lim_{n \to \infty} \left(\sqrt{n+6\sqrt{n}+1}-\sqrt{n}\right)\cdot \frac{\sqrt{n+6\sqrt{n}+1}+\sqrt{n}}{\sqrt{n+6\sqrt{n}+1}+\sqrt{n}}=\\[16pt] &=\lim_{n \to \infty} \frac{n+6\sqrt{n}+1-n}{\sqrt{n+6\sqrt{n}+1}+\sqrt{n}}=\\[16pt] &=\lim_{n \to \infty} \frac{6\sqrt{n}+1}{\sqrt{n+6\sqrt{n}+1}+\sqrt{n}}\frac{:\sqrt{n}}{:\sqrt{n}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{6\sqrt{\dfrac{n}{n}}+\sqrt{\dfrac{1}{n}}}{\sqrt{\dfrac{n}{n}+6\sqrt{\dfrac{n}{n^2}}+\dfrac{1}{n}}+\sqrt{\dfrac{n}{n}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{6+\sqrt{\dfrac{1}{n}}}{\sqrt{1+6\sqrt{\dfrac{1}{n}}+\dfrac{1}{n}}+1}=\\[16pt] &= \frac{6+\sqrt{0}}{\sqrt{1+0+0}+1}=\frac{6}{2}=3 \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \frac{\sqrt{1+2+3+...+n}}{n}$$, W liczniku pod pierwiastkiem mamy sumÄ ciÄgu arytmetycznego, zatem: $\begin{split} &\lim_{n \to \infty} \frac{\sqrt{1+2+3+...+n}}{n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\sqrt{\dfrac{1+n}{2}\cdot n}}{n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\sqrt{\dfrac{n+n^2}{2}}}{n}\cdot \dfrac{\dfrac{1}{n}}{\dfrac{1}{n}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\sqrt{\dfrac{n+n^2}{2n^2}}}{1}=\\[16pt] &=\lim_{n \to \infty} \sqrt{\frac{1}{2n}+\frac{1}{2}}=\\[16pt] &=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \frac{{\sqrt{n^2+\sqrt{n+1}}}-\sqrt{n^2-\sqrt{n-1}}}{\sqrt{n+1}-\sqrt{n}}$$, $\begin{split} &\lim_{n \to \infty} \frac{{\sqrt{n^2+\sqrt{n+1}}}-\sqrt{n^2-\sqrt{n-1}}}{\sqrt{n+1}-\sqrt{n}}=\\[16pt] &=\lim_{n \to \infty} \frac{{\sqrt{n^2+\sqrt{n+1}}}-\sqrt{n^2-\sqrt{n-1}}}{\sqrt{n+1}-\sqrt{n}}\cdot \frac{({\sqrt{n^2+\sqrt{n+1}}}+\sqrt{n^2-\sqrt{n-1}})(\sqrt{n+1}+\sqrt{n})}{({\sqrt{n^2+\sqrt{n+1}}}+\sqrt{n^2-\sqrt{n-1}})(\sqrt{n+1}+\sqrt{n})} =\\[16pt] &=\lim_{n \to \infty} \frac{(n^2+\sqrt{n+1}-n^2+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})}{\left(\sqrt{n^2+\sqrt{n+1}}+\sqrt{n^2-\sqrt{n-1}}\right)(n+1-n)}=\\[16pt] &=\lim_{n \to \infty} \frac{(\sqrt{n+1}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})}{\left(\sqrt{n^2+\sqrt{n+1}}+\sqrt{n^2-\sqrt{n-1}}\right)}=\\[16pt] &=\lim_{n \to \infty} \frac{(n+1+\sqrt{n^2+n}+\sqrt{n^2-1}+\sqrt{n^2-n})}{\left(\sqrt{n^2+\sqrt{n+1}}+\sqrt{n^2-\sqrt{n-1}}\right)}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \frac{1+\frac{1}{n}+\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}{\sqrt{1+\sqrt{\frac{1}{n^3}+\frac{1}{n^4}}}+\sqrt{1-\sqrt{\frac{1}{n^3}-\frac{1}{n^4}}}}=\\[16pt] &=\frac{1+0+1+1+1}{\sqrt{1}+\sqrt{1}}=\frac{4}{2}=2 \end{split}$, Count limit of sequence \(\lim_{n \to \infty} \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)! 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