Example. Compute the limit

$$
\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}
$$
Solution. Here $(2 n+1)^{2}-(n+1)^{2}=3 n^{2}+2 n-$ is a polynomial of the second degree (an infinitely large sequence of order $n^{2}$) and $n^{2}+n+1$ is a polynomial of the second degree (an infinitely large sequence of order $n^{2}$).

1. Factor out $n^{2}$ in the numerator, we get

$$
(2 n+1)^{2}-(n+1)^{2}=n^{2}\left(3+\frac{2}{n}\right)
$$

2. Factor out $n^{2}$ in the denominator, we get

$$
n^{2}+n+1=n^{2}\left(1+\frac{1}{n}+\frac{1}{n^{2}}\right)
$$

3. We have

$$
\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}=\lim _{n \rightarrow \infty} \frac{n^{2}(3+2 / n)}{n^{2}\left(1+1 / n+1 / n^{2}\right)}
$$

4. Canceling $n^{2}$ and using the theorem on the limit of a quotient, we get

$$
\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}=\frac{\lim _{n \rightarrow \infty}(3+2 / n)}{\lim _{n \rightarrow \infty}\left(1+1 / n+1 / n^{2}\right)}=3
$$

Answer. $\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}=3$.

Conditions of the Problems. Calculate the limit.

1. $\lim _{n \rightarrow \infty} \frac{(5-n)^{2}+(5+n)^{2}}{(5-n)^{2}-(5+n)^{2}}$. 2. $\quad \lim _{n \rightarrow \infty} \frac{(4-n)^{3}-(2-n)^{3}}{(1-n)^{2}-(2+n)^{4}}$.
2. $\lim _{n \rightarrow \infty} \frac{(3-n)^{3}-(2-n)^{3}}{(1-n)^{3}-(1+n)^{3}}$.
3. $\lim _{n \rightarrow \infty} \frac{(2-n)^{2}-(1+n)^{2}}{(1+n)^{2}-(2-n)^{2}}$.
4. $\lim _{n \rightarrow \infty} \frac{(3+n)^{2}-(2+n)^{2}}{(2+n)^{2}-(1-n)^{2}}$.
5. $\lim _{n \rightarrow \infty} \frac{(n+2)^{3}-(n+2)^{2}}{(n-2)^{3}-(n+2)^{3}}$.
6. $\lim _{n \rightarrow \infty} \frac{(1+3 n)^{3}-27 n^{3}}{(1+4 n)^{2}+2 n^{2}}$.
7. $\lim _{n \rightarrow \infty} \frac{(3-2 n)^{2}}{(n-3)^{3}-(n+3)^{3}}$.
8. $\lim _{n \rightarrow \infty} \frac{(2+n)^{3}}{(n+2)^{2}-(n+1)^{3}}$.
9. $\lim _{n \rightarrow \infty} \frac{(n+2)^{2}-(n+5)^{3}}{(3-n)^{3}}$.

Answers. 1. $-\infty$. 2.0.3.0. 4. $-1.5 .1 / 3.6 .-\infty .7 .9 .8 .-2 / 9$. 9. -1.10 .1 .

## 3.3. Calculation of $\lim _{n \rightarrow \infty}[f(n) / g(n)]$

Problem Statement. Calculate the limit

$$
\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}
$$

where $f(n)$ is an infinitely large sequence of order $n^{\alpha}$ and $g(n)$ is an infinitely large sequence of order $n^{\beta}$ $(\alpha, \beta \in \mathbb{R})$.

Plan of Solution.

1. Factor out $n^{\alpha}$ in the numerator, we get $f(n)=n^{\alpha} \varphi(n)$, where $\lim _{n \rightarrow \infty} \varphi(n)=a, a \neq 0$.
2. Factor out $n^{\beta}$ in the denominator, we get $g(n)=n^{\beta} \psi(n)$, where $\lim _{n \rightarrow \infty} \psi(n)=b, b \neq 0$.
3. We have

$$
\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=\lim _{n \rightarrow \infty} \frac{n^{\alpha} \varphi(n)}{n^{\beta} \psi(n)}
$$

4. We get:

if $\alpha>\beta$, then $\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=\infty$;

if $\alpha<\beta$, then $\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=0$;

if $\alpha=\beta$, then by the theorem on the limit of a quotient

$$
\lim _{n \rightarrow \infty} \frac{f(n)}{g(n)}=\frac{\lim _{n \rightarrow \infty} \varphi(n)}{\lim _{n \rightarrow \infty} \psi(n)}=\frac{a}{b}
$$