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Step 1: Choose a series of x-values that are very close to the stated x-value, coming from the left of the number line. 1 is an upper bound and the supremum. Approaching Infinity. A sequence {zn} is a Cauchy sequence iﬀ for each ε>0, there is Nε such that m,n ≥ Nε implies |zm −zn|≤ε (in short, lim m,n→∞ |zn − zm| = 0). For a sequence of real numbers, the largest accumulation point is called the limit superior and denoted by lim sup or . A sequence is bounded below if and only if inf L > −∞. Examples In this case, we say that x 0 is the limit of the sequence and write x n := x 0 . R with the usual metric Sets sometimes contain their limit points and sometimes do not. $\lim_{n \to \infty} \frac{1}{n^2}+\frac{2}{n^3}-\frac{100}{n^6}=0+0-0=0$, 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 $$\lim_{n \to \infty} \sqrt{2}\cdot \sqrt[4]{2}\cdot \sqrt[8]{2}\cdot ...\cdot \sqrt[2^n]{2}$$, $\begin{split} &\lim_{n \to \infty} \sqrt{2}\cdot \sqrt[4]{2}\cdot \sqrt[8]{2}\cdot ...\cdot \sqrt[2^n]{2}=\\[16pt] &=\lim_{n \to \infty} 2^\tfrac{1}{2}\cdot 2^\tfrac{1}{4}\cdot ...\cdot 2^\tfrac{1}{2^n}=\\[16pt] &=\lim_{n \to \infty} 2^{\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+...+\tfrac{1}{2^n}}=\\[16pt] &= 2^{\tfrac{\tfrac{1}{2}}{1-\tfrac{1}{2}}}=\\[16pt] &= 2^{\tfrac{1}{2}\cdot \tfrac{2}{1}}=\\[16pt] &= 2^1=2 \end{split}$, 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)! Infinity is a very special idea. So by de nition, y 2S. Example of a Loop Sequence. In other words, a point x of a topological space X is said to be the limit point of a subset A of X if for every open set U containing x we have Example: Find the limit of the sequence {n-n 3. Let {f n} be the sequence of functions on R deﬁned by f n(x) = nx. It is equivalent to say that for every neighbourhood {\displaystyle V} of {\displaystyle x} and every {\displaystyle n_{0}\in \mathbb {N} }, there is some {\displaystyle n\geq n_{0}} such that {\displaystyle x_{n}\in V}. For example, if a sequence tends to inﬁnity or to minus inﬁnity then it is divergent. Limit points are also called accumulation points. In the real numbers every Cauchy sequence converges to some limit. A point x2R is a limit point of Aif every -neighborhood V (x) of xintersects A at some point other than x, i.e. (2) There is a sequence (xn) in A with xn ̸= c such that limn!1 xn = c but the sequence (f(xn)) does not converge. 2. If {\displaystyle X} is a metric space or a first-countable space (or, more generally, a Fréchet–Urysohn space), then {\displaystyle x} is cluster point of {\displaystyle (x_{n})_{n\in \mathbb {N} }} if and only if {\displaystyle x} is a limit of some subsequence of {\displaystyle (x_{n})_{n\in \mathbb {N} }}. If for every \varepsilon > 0{\text{ }}\exists m \in \mathbb{N} such that {u_n} \in \left( {l – \varepsilon ,l + \varepsilon } \right), \forall n \geqslant m or equivalently \left| {{u_n} – l} \right| < \varepsilon  \forall n \geqslant m, then l is a limit point of the sequence u .In such a case it can be easily seen that l is the only limit point of the sequence. Conclusively, it follows that the limit points of a sequence u are either the points or the limit points of the set R\left\{ u \right\}. We then say that zero is the limit (or sometimes the limiting value) of the sequence and write, lim n → ∞an = lim n → ∞n + 1 n2 = 0. lim n → ∞ a n = lim n → ∞ n + 1 n … We have to verify the deﬁnition above with ‘ = 0. Examples. If we look at the subsequence of odd terms we have that its limit is -1… For example, x n = ( 1) n2 n n!. But this distinction is not necessary. Now pick the tolerance 0 Should it be necessary that sequence values are never equal to its limit? A positive number \eta  is said to be arbitrarily small if given any \varepsilon > 0, \eta  may be chosen such that 0 < \eta < \varepsilon . Deﬁnition. 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 . Lemma. As in the case of sets of real numbers, limit points of a sequence may also be called accumulation, cluster or condensation points. Example 4: The set of limit points of a bounded sequence u is bounded. f(x) = x 2 as x → 3 from below. The set L is a closed set, i.e. Example: A bounded closed subset of is sequentially compact, by Heine-Borel Theorem. But S is just one point, a, so we have y = a and we have shown a1 n k!a too. Example 1: Limit Points (a)Let c 0 it is implied that \varepsilon  may be howsoever small positive number. Definition. 1 Limits of Sequences De nition 1 (Sequence) Let Xbe a set. 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. Sequence Diagram Example of a School Management System . For anyϵ >0, there are at most … Deﬁnition A sequence which has a limit is said to be convergent. In sequences (unlike sets) an infinitely repeating term counts as infinitely many terms. A point x2R is a limit point of Aif every -neighborhood V (x) of xintersects A at some point other than x, i.e. Evidently, if l = {u_n} for infinitely many values of n then l is a limit point of the sequence u. Exercises on Limit Points. An example of such a sequence is the sequence $u_n = \frac{n}{2}(1+(-1)^n),$ whose initial values are $0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 6, \dots$ \((u_n)$$ is an unbounded sequence whose unique limit point is $$0$$. The Limit of a Sequence The concept of determining if sequence converges or diverges. The set of limit points of (c;d) is [c;d]. Introduction In order to make us understand the information more on approaches of a given real sequence an n 1 , we give two definitions, thier names are upper limit and lower limit.It is fundamental but important tools in analysis. The higher is, the smaller is and the closer it gets to .Therefore, intuitively, the limit of the sequence should be : It is straightforward to prove that is indeed a limit … Example The sequence 1 n ∈N is convergent with limit 0. For example f1=n : n 2Ngconverges in R1 and diverges in (0;1). Please Subscribe here, thank you!!! R 2 with the usual metric }=\$16pt] &=\lim_{n \to \infty} \frac{(n+1)!\cdot (n+2+1)}{(n+1)!\cdot (n+2-1)}=\\[16pt] &=\lim_{n \to \infty} \frac{(n+1)!\cdot (n+3)}{(n+1)!\cdot (n+1)}=\\[16pt] &=\lim_{n \to \infty} \frac{n+3}{n+1}\cdot \frac{\frac{1}{n}}{\frac{1}{n}}=\\[16pt] &=\lim_{n \to \infty} \frac{1+\frac{3}{n}}{1+\frac{1}{n}}=\\[16pt] &=\frac{1}{1}=1 \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \frac{7^n+5^n}{5^n+3^n}$$, $\begin{split} \lim_{n \to \infty} \frac{7^n+5^n}{5^n+3^n}&=\lim_{n \to \infty} \frac{7^n\left(1+\left(\dfrac{5}{7}\right)^n\right)}{5^n\left(1+\left(\dfrac{3}{5}\right)^n\right)}=\\[16pt] &=\lim_{n \to \infty} \left(\frac{7}{5}\right)^n=\infty \end{split}$, Count limit of sequence $$\lim_{n \to \infty} n(\ln (n+1)-\ln n)$$, $\begin{split} &\lim_{n \to \infty} n\left(\ln (n+1)-\ln n\right)=\\[16pt] &=\lim_{n \to \infty} n\left(\ln \frac{n+1}{n}\right)=\\[16pt] &=\lim_{n \to \infty} \ln \left(\frac{n+1}{n}\right)^n=\\[16pt] &=\lim_{n \to \infty} \ln \left(1+\frac{1}{n}\right)^n=\\[16pt] &=\lim_{n \to \infty} \ln e=1 \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \frac{\log_2(n+1)}{\log_3(n+1)}$$, $\begin{split} &\lim_{n \to \infty} \frac{\log_2(n+1)}{\log_3(n+1)}=\\[16pt] &=\lim_{n \to \infty} \frac{\log_2(n+1)}{\frac{\log_2(n+1)}{\log_23}}=\\[16pt] &=\lim_{n \to \infty} \log_23=\\[16pt] &=\log_23 \end{split}$, Count limit of sequence $$\lim_{n \to \infty} (1+2^n-3^n)$$, $\begin{split} &\lim_{n \to \infty} (1+2^n-3^n)=\\[16pt] &=\lim_{n \to \infty} 3^n\left(\frac{1}{3^n}+\left(\frac{2}{3}\right)^n-1\right)=\\[16pt] &=-\lim_{n \to \infty} 3^n=-\infty \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \left(\frac{n+5}{n}\right)^n$$, $\begin{split} &\lim_{n \to \infty} \left(\frac{n+5}{n}\right)^n=\\[16pt] &=\lim_{n \to \infty} \left(1+\frac{5}{n}\right)^n=\\[16pt] &=\lim_{n \to \infty}\left(1+\frac{1}{\frac{n}{5}}\right)^n=\\[16pt] &=\lim_{n \to \infty} \left[\left(1+\frac{1}{\frac{n}{5}}\right)^\dfrac{n}{5}\right]^5=\\[16pt] &=e^5 \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \left(1-\frac{1}{n^2}\right)^n$$, $\begin{split} &\lim_{n \to \infty} \left(1-\frac{1}{n^2}\right)^n=\\[16pt] &=\lim_{n \to \infty} \left(\left(1-\frac{1}{n^2}\right)^{n^2}\right)^{\frac{1}{n}}=\\[16pt] &=e^0=1 \end{split}$, Count limit of sequence $$\lim_{n \to \infty} \left(\frac{n^2+6}{n^2}\right)^{n^2}$$, $\begin{split} &\lim_{n \to \infty} \left(\frac{n^2+6}{n^2}\right)^{n^2}=\\[6pt] &=\lim_{n \to \infty} \left(1+\frac{6}{n^2}\right)^{n^2}=\\[6pt] &=\lim_{n \to \infty} \left[\left(1+\frac{6}{n^2}\right)^{\dfrac{n^2}{6}}\right]^6=\\[6pt] &=e^6 \end{split}$, Count limit of function $$\lim_{x \to {-3}} (x^2+3x+7)$$, $\lim_{x \to {-3}} (x^2+3x+7)=(-3)^2+3\cdot (-3)+7=7$, Count limit of function $$\lim_{x \to 4}\frac{\sqrt{x^2-16}}{4x+2}$$, $\lim_{x \to 4}\frac{\sqrt{x^2-16}}{4x+2} =\frac{\sqrt{4^2-16}}{4\cdot 4+2}=\frac{0}{18}=0$, Count limit of function $$\lim_{x \to 2}\frac{x^2-6x+9}{x^2-9}$$, $\begin{split} &\lim_{x \to 3}\frac{x^2-6x+9}{x^2-9} =\\[15pt] &=\lim_{x \to 3}\frac{(x-3)^2}{(x-3)(x+3)}=\\[15pt] &=\lim_{x \to 3}\frac{x-3}{x+3}=\\[15pt] &=\frac{0}{6}=0 \end{split}$, Count limit of function $$\lim_{x \to 1}\frac{1-x^2}{\left(1-\sqrt{x}\right)}$$, $\begin{split} &\lim_{x \to 1}\frac{1-x^2}{\left(1-\sqrt{x}\right)} =\\[16pt] &=\lim_{x \to 1}\frac{(1-x)(1+x)}{\left(1-\sqrt{x}\right)} =\\[16pt] &=\lim_{x \to 1}\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)(1+x)}{\left(1-\sqrt{x}\right)} =\\[16pt] &=\lim_{x \to 1} \left(1+\sqrt{x}\right)(1+x)=\\[16pt] &=2\cdot 2=4 \end{split}$, Count limit of function $$\lim_{z \to -2} \frac{z^3+4z^2+4z}{z^2-z-6}$$, $\begin{split} &\lim_{z \to -2}\frac{z(z^2+4z+4)}{(z+2)(z-3)}=\\[16pt] &=\lim_{z \to -2}\frac{z(z+2)^2}{(z+2)(z-3)}=\\[16pt] &=\lim_{z \to -2}\frac{z(z+2)}{z-3} =\\[16pt] &=\frac{0}{-5}=0 \end{split}$, Count limit of function $$\lim_{n \to \infty} \frac{2x^2-1}{7x^2+2x}$$, $\begin{split} &\lim_{n \to \infty} \frac{2x^2-1}{7x^2+2x}=\\[16pt] &=\lim_{n \to \infty} \frac{2x^2-1}{7x^2+2x}\cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\\[16pt] &=\lim_{n \to \infty} \frac{2-\dfrac{1}{x^2}}{7+\dfrac{2}{x}}=\\[16pt] &=\frac{2}{7} \end{split}$, Count limit of function $$\lim_{n \to -\infty}\frac{1+\sqrt{2x^2-1}}{x}$$, $\begin{split} &\lim_{n \to -\infty}\frac{1+\sqrt{2x^2-1}}{x}=\\[16pt] &=\lim_{n \to -\infty}\frac{1+\sqrt{2x^2-1}}{x}\cdot \frac{\frac{1}{|x|}}{\frac{1}{|x|}}=\\[16pt] &=\lim_{n \to -\infty}\dfrac{\dfrac{1}{|x|}+\sqrt{\dfrac{2x^2}{|x^2|}-\dfrac{1}{|x^2|}}}{\dfrac{x}{|x|}}=\\[16pt] &=\lim_{n \to -\infty}\dfrac{\dfrac{1}{|x|}+\sqrt{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}}{-1}=\\[16pt] &=\lim_{n \to -\infty}\dfrac{\dfrac{1}{|x|}+\sqrt{2-\dfrac{1}{x^2}}}{-1}=\\[16pt] &=\dfrac{0+\sqrt{2-0}}{-1}=\\[16pt] &=-\sqrt{2} \end{split}$, Count limit of function $$\lim_{x \to 2}\left(\frac{1}{x-2}-\frac{4}{x^2-4}\right)$$, $\begin{split} &\lim_{x \to 2}\left(\frac{1}{x-2}-\frac{4}{x^2-4}\right)=\\[16pt] &=\lim_{x \to 2}\left(\frac{x+2}{(x-2)(x+2)}-\frac{4}{(x-2)(x+2)}\right)=\\[16pt] &=\lim_{x \to 2}\frac{(x-2)}{(x-2)(x+2)}=\\[16pt] &=\lim_{x \to 2}\frac{1}{x+2}=\frac{1}{4} \end{split}$, Count limit of function $$\lim_{x \to 2} \frac{x^4-8x^2+16}{(x-2)(x-3)}$$, $\begin{split} &\lim_{x \to 2} \frac{x^4-8x^2+16}{(x-2)(x-3)}=\\[16pt] &=\lim_{x \to 2}\frac{(x^2-4)^2}{(x-2)(x-3)}=\\[16pt] &=\lim_{x \to 2}\frac{(x-2)^2(x+2)^2}{(x-2)(x-3)}=\\[16pt] &=\lim_{x \to 2}\frac{(x-2)(x+2)^2}{x-3}=\\[16pt] &=\frac{0}{-1}=0 \end{split}$, Count limit of function $$\lim_{x \to \infty} \frac{\sqrt{1+x}+3x}{\sqrt{1+x^2}}$$, $\begin{split} &\lim_{x \to \infty} \frac{\sqrt{1+x}+3x}{\sqrt{1+x^2}}=\\[16pt] &=\lim_{x \to \infty} \frac{\sqrt{1+x}+3x}{\sqrt{1+x^2}}\cdot \frac{\frac{1}{x}}{\frac{1}{x}}=\\[16pt] &=\lim_{x \to \infty} \frac{\sqrt{\dfrac{1}{x^2}+\dfrac{x}{x^2}}+\dfrac{3x}{x}}{\sqrt{\dfrac{1}{x^2}+\dfrac{x^2}{x^2}}}=\\[16pt] &=\lim_{x \to \infty} \frac{\sqrt{\dfrac{1}{x^2}+\dfrac{1}{x}}+3}{\sqrt{\dfrac{1}{x^2}+1}}=\\[16pt] &=\frac{3}{\sqrt{1}}=3 \end{split}$, Count limit of function $$\lim_{x \to -1} \frac{x^4+3x^2-4}{x+1}$$, $\begin{split} &\lim_{x \to -1} \frac{x^4+3x^2-4}{x+1}=\\[16pt] &=\lim_{x \to -1}\frac{(x^2+4)(x^2-1)}{x+1}=\\[16pt] &=\lim_{x \to -1}\frac{(x^2+4)(x-1)(x+1)}{x+1}=\\[16pt] &=\lim_{x \to -1}(x^2+4)(x-1)=\\[16pt] &=5\cdot (-2)=-10 \end{split}$, Count limit of function $$\lim_{x \to 0} \frac{\sin 2x}{x}$$, $\lim_{x \to 0} \frac{\sin 2x}{x}=\lim_{x \to 0}\frac{2\sin 2x}{2x}=2\lim_{x \to 0}\frac{\sin 2x}{2x}=2$, Count limit of function $$\lim_{x \to -1}\frac{\sin (x+1)}{1-x^2}$$, $\begin{split} &\lim_{x \to -1}\frac{\sin (x+1)}{1-x^2}=\\[16pt] &=\lim_{x \to -1}\frac{\sin (x+1)}{(1-x)(1+x)}=\\[16pt] &=\lim_{x \to -1}\frac{\sin (x+1)}{1+x}\cdot \lim_{x \to -1}\frac{1}{1-x}=\\[16pt] &=1\cdot \frac{1}{2}=\frac{1}{2} \end{split}$, Count limit of function $$\lim_{x \to \dfrac{\pi }{4}}\frac{\sqrt{\sin x}-\sqrt{\cos x}}{\sin x-\cos x}$$, $\begin{split} &\lim_{x \to \dfrac{\pi }{4}}\frac{\sqrt{\sin x}-\sqrt{\cos x}}{\sin x-\cos x} =\\[16pt] &=\lim_{x \to \dfrac{\pi }{4}}\frac{(\sqrt{\sin x}-\sqrt{\cos x})}{(\sqrt{\sin x}-\sqrt{\cos x})(\sqrt{\sin x}+\sqrt{\cos x})}=\\[16pt] &=\lim_{x \to \dfrac{\pi }{4}}\frac{1}{\sqrt{\sin x}+\sqrt{\cos x}}=\\[16pt] &=\frac{1}{\sqrt{\frac{\sqrt{2}}{2}}+\sqrt{\frac{\sqrt{2}}{2}}}=\frac{1}{\sqrt{2\sqrt{2}}} \end{split}$, © 2010-2015 Matemaks MichaÅ BudzyÅski |. 267 Let S= [ a ; b ] and x2 [ a ; b ] 0 example 266 bare. 3 from below values are never equal to its limit an output in x examples on this page are. 1 n+ 2 does not converge to a real limit ( but not differentiable limit. 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