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:

$ \lim \dfrac{2n^3-n^2+1}{n^3+1}=2 $
$ \lim \dfrac{-n^7-n^6+1}{n+2n^7}=-\dfrac{1}{2} $
$ \lim \dfrac{(n+1)(n^2-3n+5)}{n^3-2n(n^2+1)+2}=-1 $
$ \lim \dfrac{n\sqrt{n}+n^2-1}{2n^2+1}=\dfrac{1}{2} $
$ \lim \dfrac{n+\sqrt{n^2+1}}{3n-1}=\dfrac{2}{3} $
$ \lim \dfrac{n+\sqrt{n^2+2n+1}}{2-\sqrt{4n^2+1}}=-1 $
$ \lim \dfrac{2n+\sqrt{n^3+2n^2+1}}{n\sqrt{3n+2}-1}=\dfrac{1}{\sqrt{3}} $
$ \lim \dfrac{\sqrt[3]{-27n^6+2n+1}}{4n^2+4n+1}=-\dfrac{3}{4} $
$ \lim \sqrt{\dfrac{3n^2+2n-1}{n^2+5n}}=\sqrt{3} $
$ \lim \left(\dfrac{2n^2}{n^2+3n+1}-\dfrac{2n}{3n+1}\right) $
$ \lim \dfrac{n+1}{n^3+1}=0 $
$ \lim \dfrac{11n^2-2n+1}{n^3+n^2+1}=0 $
$ \lim \dfrac{(2n+1)(n-5)+n^2+1}{n^3+n^2}=0 $
$ \lim \left(\dfrac{2n}{3n^2+1}-\dfrac{n}{3n^2+1}\right) $
$ \lim \dfrac{n+\sqrt{n^3+1}}{2n^2+\sqrt{n}-1}=0 $
$ \lim \dfrac{2}{\sqrt{n^2+1}-n} $
$ \lim \dfrac{1}{\sqrt{n}\left(\sqrt{n+2}-\sqrt{n+1}\right)} $
$ \lim \dfrac{3}{\sqrt{4n^2+1}-2n+1} $
$ \lim \dfrac{2^n+3^n}{5\cdot3^n+2^n} $
$ \lim \dfrac{2^n-1}{3^n+2^{n+1}} $
$ \lim \dfrac{2^n-3^n+5^{n+2}}{5^n+3^{n+1}} $
$ \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:

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

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

$ \lim (n\sqrt{n}+n-3)=+\infty $
$ \lim (\sqrt{n^2+n+1}+3n)=+\infty $
$ \lim (\sqrt{n^2+1}-3n)=-\infty $
$ \lim (\sqrt{n+1}+\sqrt{n+2})=+\infty $
$ \lim (\sqrt{n+1}-\sqrt{n+2})=0 $
$ \lim \left(\dfrac{1}{n-\sqrt{n^2+1}}-\dfrac{1}{n+\sqrt{n^2+1}}\right)$
$ \lim (\sqrt{n^2+n+1}-n)$
$ \lim (\sqrt{n^2+n+1}-\sqrt{n^2-n+1})$
$ \lim \dfrac{3n+2}{\sqrt{n^2+3}-\sqrt{n^2+1}}$
$ \lim \dfrac{\sqrt{n^2+n}-n}{\sqrt{n^2+1}-\sqrt{n^2+2n}}$
$ \lim (\sqrt[3]{n^3+1}-n)$
$ \lim (2n+1+\sqrt[3]{1-8n^3})$

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

$ \lim n(\sqrt{n^2+1}-n)=+\infty $
$ \lim \sqrt{n+1}(\sqrt{n+2}-\sqrt{n}) $
$ \lim n^2(\sqrt{3n^4+5}-\sqrt{3n^4+2}) $
$ \lim \dfrac{n(n+\sqrt{n-n^3})}{n-\sqrt{n^2+4n}} $
$ \lim (\sqrt{n^2+1}-\sqrt[3]{n^3+n})$
$ \lim (\sqrt{n^2+n+1}+\sqrt[3]{1-n^3})$
$ \lim (2n-\sqrt{9n^2+n}+\sqrt{n^2+2n}) $
$ \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 \[ 0.777…=\frac{u_1}{1-q}=\frac{\frac{7}{10}}{1-\frac{1}{10}}=\frac{7}{9} \]

Limit of a Sequence