# Limit of a sequence

## 1. Basic keywords

### Some basic limits:

• $\lim c=c$
• $\lim \frac{1}{n}=0$
• $\lim \frac{c}{n^k}=0$ for some positive integers $k$
• $\lim \frac{1}{\sqrt{n}}=0$
• $\lim \frac{1}{\sqrt[k]{n}}=0$ for some positive integers $k$
• $\lim q^n=0$ for $|q|<1$
• $\lim n=+\infty$
• $\lim n^k=+\infty$ for some positive integers $k$
• $\lim \sqrt{n}=+\infty$
• $\lim \sqrt[k]{n}=+\infty$ for some positive integers $k$
• $\lim q^n=+\infty$ for $q>1.$

### Squeeze Law

Let three sequences $(u_n),(v_n)$ and $(w_n)$ such that $u_n\le v_n\le w_n$ and $\lim u_n=\lim w_n=L ,$ then $\lim v_n=L$
Specially, if $|u_n|\le v_n$ and $\lim v_n=0$ then $\lim u_n=0.$

### Basic Limit Law

Finite limits: If $\lim u_n=A$ and $\lim v_n=B$ then

• $\lim (u_n\pm v_n)=\lim A\pm \lim B$
• $\lim (c\cdot u_n)= c\cdot A$
• $\lim (u_n\cdot v_n)=AB$
• $\lim \frac{u_n}{v_n}=\frac{A}{B}$ if $b\ne 0$

### Infinite Limits:

• $\lim (u_n\cdot v_n)$
• $\lim \frac{u_n}{v_n}$

## 2. Example

Example 1. Find the limit of the following sequence, or determine that the limit does not exist: $$\lim \frac{3n^3-1}{2n^3-1}$$
Hint. Divide numerator and denominator by $n^3,$ we get \begin{align*}
\lim \frac{3n^3-1}{2n^3-1} &=\lim\frac{3-\frac{1}{n^3}}{2+\frac{1}{n^3}}\\
&=\frac{3}{2}
\end{align*}
So $\lim \frac{3n^3-1}{2n^3-1} =\frac{3}{2}.$

Example 2.  Find the limit of the following sequences if it exists:

1. $\lim \dfrac{2n^3-n^2+1}{n^3+1}=2$
2. $\lim \dfrac{-n^7-n^6+1}{n+2n^7}=-\dfrac{1}{2}$
3. $\lim \dfrac{(n+1)(n^2-3n+5)}{n^3-2n(n^2+1)+2}=-1$
4. $\lim \dfrac{n\sqrt{n}+n^2-1}{2n^2+1}=\dfrac{1}{2}$
5. $\lim \dfrac{n+\sqrt{n^2+1}}{3n-1}=\dfrac{2}{3}$
6. $\lim \dfrac{n+\sqrt{n^2+2n+1}}{2-\sqrt{4n^2+1}}=-1$
7. $\lim \dfrac{2n+\sqrt{n^3+2n^2+1}}{n\sqrt{3n+2}-1}=\dfrac{1}{\sqrt{3}}$
8. $\lim \dfrac{\sqrt[3]{-27n^6+2n+1}}{4n^2+4n+1}=-\dfrac{3}{4}$
9. $\lim \sqrt{\dfrac{3n^2+2n-1}{n^2+5n}}=\sqrt{3}$
10. $\lim \left(\dfrac{2n^2}{n^2+3n+1}-\dfrac{2n}{3n+1}\right)$
11. $\lim \dfrac{n+1}{n^3+1}=0$
12. $\lim \dfrac{11n^2-2n+1}{n^3+n^2+1}=0$
13. $\lim \dfrac{(2n+1)(n-5)+n^2+1}{n^3+n^2}=0$
14. $\lim \left(\dfrac{2n}{3n^2+1}-\dfrac{n}{3n^2+1}\right)$
15. $\lim \dfrac{n+\sqrt{n^3+1}}{2n^2+\sqrt{n}-1}=0$
16. $\lim \dfrac{2}{\sqrt{n^2+1}-n}$
17. $\lim \dfrac{1}{\sqrt{n}\left(\sqrt{n+2}-\sqrt{n+1}\right)}$
18. $\lim \dfrac{3}{\sqrt{4n^2+1}-2n+1}$
19. $\lim \dfrac{2^n+3^n}{5\cdot3^n+2^n}$
20. $\lim \dfrac{2^n-1}{3^n+2^{n+1}}$
21. $\lim \dfrac{2^n-3^n+5^{n+2}}{5^n+3^{n+1}}$
22. $\lim \dfrac{(-2)^n-5^{n+1}}{5^{n-1}+3^{n+1}}$

Example 3. Find the limit of the following sequences if it exists:

1. $\lim (n^2+n-3)=+\infty$
2. $\lim (2n^2-n^3+4)=-\infty$
3. $\lim (n\sqrt{n}+3n-1)=+\infty$
4. $\lim \dfrac{2n^3-n^2+1}{n^2+1}=+\infty$
5. $\lim \dfrac{11n^4+1}{-5n^2+n+1}=-\infty$
6. $\lim \dfrac{2n\sqrt{n}-3}{n+\sqrt{n}-1}$
7. $\lim \dfrac{3^n+2^n}{2^{n+1}-1}$

Example 4. Find the limit of the following sequences if it exists:

1. $\lim (n\sqrt{n}+n-3)=+\infty$
2. $\lim (\sqrt{n^2+n+1}+3n)=+\infty$
3. $\lim (\sqrt{n^2+1}-3n)=-\infty$
4. $\lim (\sqrt{n+1}+\sqrt{n+2})=+\infty$
5. $\lim (\sqrt{n+1}-\sqrt{n+2})=0$
6. $\lim \left(\dfrac{1}{n-\sqrt{n^2+1}}-\dfrac{1}{n+\sqrt{n^2+1}}\right)$
7. $\lim (\sqrt{n^2+n+1}-n)$
8. $\lim (\sqrt{n^2+n+1}-\sqrt{n^2-n+1})$
9. $\lim \dfrac{3n+2}{\sqrt{n^2+3}-\sqrt{n^2+1}}$
10. $\lim \dfrac{\sqrt{n^2+n}-n}{\sqrt{n^2+1}-\sqrt{n^2+2n}}$
11. $\lim (\sqrt[3]{n^3+1}-n)$
12. $\lim (2n+1+\sqrt[3]{1-8n^3})$

Example 5. Find the limit of the following sequences if it exists:

1. $\lim n(\sqrt{n^2+1}-n)=+\infty$
2. $\lim \sqrt{n+1}(\sqrt{n+2}-\sqrt{n})$
3. $\lim n^2(\sqrt{3n^4+5}-\sqrt{3n^4+2})$
4. $\lim \dfrac{n(n+\sqrt{n-n^3})}{n-\sqrt{n^2+4n}}$
5. $\lim (\sqrt{n^2+1}-\sqrt[3]{n^3+n})$
6. $\lim (\sqrt{n^2+n+1}+\sqrt[3]{1-n^3})$
7. $\lim (2n-\sqrt{9n^2+n}+\sqrt{n^2+2n})$
8. $\lim \left(\sqrt{n^2+2n}+2\sqrt[3]{n^2-8n^3}+\sqrt{n^2+1}\right)$

Example 6. Find the limit of the following sequence $\lim \frac{1+2+3+\cdots+n}{1+n^3}$

Example 7. Find the limit of the following sequences: $u_n=\dfrac{\sin(2n+1)}{3^n},v_n=\dfrac{(-1)^n}{2n+3}$
Hint. For all $n$ we have $\left|\frac{\sin(2n+1)}{3^n}\right|\le \frac{1}{3^n}$
and $\lim \frac{1}{3^n}=0$ and so $\lim \frac{\sin(2n+1)}{3^n}=0.$

Example 8. Express the repeating decimal $0.777…$ as a fraction.
Hint. We have \begin{align*} 0.777…&=0.7+0.07+0.0007+\cdots\\
&=\frac{7}{10}+\frac{7}{100}+\frac{7}{1000}+\cdots \end{align*}
This is the sum of an geometric sequence with $u_1=\frac{7}{10}$ and the common ratio $q=\frac{1}{10}<1$. So $0.777…=\frac{u_1}{1-q}=\frac{\frac{7}{10}}{1-\frac{1}{10}}=\frac{7}{9}$