Example. Compute the limit

$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}
$$
Solution. The numerator $n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}$ is an infinitely large sequence of order $n^{2}$, and the denominator $(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}$ is an infinitely large sequence of order $n^{2}$.

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

$$
n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}=n^{2}\left(\frac{1}{n^{5 / 6}}+2 \sqrt[5]{1+\frac{1}{n^{10}}}\right)
$$

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

$$
(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}=n^{2}\left(1+\frac{1}{n^{3 / 4}}\right) \sqrt[3]{1-\frac{1}{n^{3}}}
$$

3. We have

$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}=\lim _{n \rightarrow \infty} \frac{n^{2}\left(1 / n^{5 / 6}+2 \sqrt[5]{1+1 / n^{10}}\right)}{n^{2}\left(1+1 / n^{3 / 4}\right) \sqrt[3]{1-1 / n^{3}}}
$$

4. Canceling $n^{2}$ and using theorems about limits, we finally get

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}=\lim _{n \rightarrow \infty} \frac{1 / n^{5 / 6}+2 \sqrt[5]{1+1 / n^{10}}}{\left(1+1 / n^{3 / 4}\right) \sqrt[3]{1-1 / n^{3}}}= \\
&=\frac{\lim _{n \rightarrow \infty}\left(1 / n^{5 / 6}+2 \sqrt[5]{1+1 / n^{10}}\right)}{\lim _{n \rightarrow \infty}\left(1+1 / n^{3 / 4}\right) \sqrt[3]{1-1 / n^{3}}}=2
\end{aligned}
$$

NOTE. In this case, the property of the root was used, according to which $\lim _{n \rightarrow \infty} \sqrt[5]{1+1 / n^{10}}=1$ and $\lim _{n \rightarrow \infty} \sqrt[3]{1-1 / n^{3}}=1$.

$$
\text { Answer. } \lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}=2
$$

Conditions of the Problems. Calculate the limit.

1. $\lim _{n \rightarrow \infty} \frac{n \sqrt[3]{3 n^{2}}+\sqrt[4]{4 n^{8}+1}}{(n+\sqrt{n}) \sqrt{7-n+n^{2}}}$.
2. $\lim _{n \rightarrow \infty} \frac{\sqrt{n-1}-\sqrt{2 n^{2}+3}}{\sqrt[3]{n^{3}+3}+\sqrt[4]{n^{5}+2}}$.
3. $\lim _{n \rightarrow \infty} \frac{\sqrt{2 n^{3}+3}-\sqrt{n+5}}{\sqrt[3]{n^{3}+2}-\sqrt{n-1}}$.
4. $\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}+3}+3 n^{3}}{\sqrt[4]{n^{12}+2 n+1}-n^{2}}$.
5. $\lim _{n \rightarrow \infty} \frac{\sqrt{3 n+2}-\sqrt[3]{125 n^{3}+n}}{\sqrt[5]{n}+n^{2}}$.
6. $\lim _{n \rightarrow \infty} \frac{n \sqrt{n}-\sqrt[3]{27 n^{6}+n^{4}}}{(n+\sqrt[4]{n}) \sqrt{4+n^{2}}}$.
7. $\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt{n^{2}+2}}{\sqrt[4]{n^{4}+1}-\sqrt[3]{n^{2}-1}}$.
8. $\lim _{n \rightarrow \infty} \frac{\sqrt{n^{5}+3}+\sqrt{n-2}}{\sqrt[4]{n^{4}+2}-\sqrt{n-2}}$.
9. $\lim _{n \rightarrow \infty} \frac{10 n^{3}-\sqrt{n^{3}+2}}{\sqrt{4 n^{6}+3}-n}$.
10. $\lim _{n \rightarrow \infty} \frac{\sqrt{n+2}-\sqrt[3]{8 n^{3}+3}}{\sqrt[4]{n+5}+n}$.

Answers. 1. $\sqrt{2}$. 2.0.3. $+\infty .4$. 3. 5.5. 6. -3. 7. -1 . 8. $+\infty$. 9.5. 10. -2 .

## 3.4. Calculation of $\lim _{n \rightarrow \infty}\left[u(n)^{v(n)}\right]$

Problem Statement. Calculate the limit of the sequence

$$
\lim _{n \rightarrow \infty}\left[u(n)^{v(n)}\right]
$$

where $\lim _{n \rightarrow \infty} u(n)=1$ and $\lim _{n \rightarrow \infty} v(n)=\infty$.

## Plan of Solution.

1. Transform the expression under the limit so that we can use the second remarkable limit, i.e., factor out the one:

$$
\lim _{n \rightarrow \infty}\left[u(n)^{v(n)}\right]=\lim _{n \rightarrow \infty}\left((1+\alpha(n))^{1 / \alpha(n)}\right)^{\alpha(n) v(n)}
$$

where $\alpha(n)=u(n)-1$ is an infinitesimal sequence as $n \rightarrow \infty$. Since $\alpha(n) \rightarrow 0$ as $n \rightarrow \infty$, we have

$$
\lim _{n \rightarrow \infty}(1+\alpha(n))^{1 / \alpha(n)}=e
$$

2. If $\lim _{n \rightarrow \infty} a_{n}=a\left(a_{n}>0, a>0\right)$ and $\lim _{n \rightarrow \infty} b_{n}=b$, then

$$
\lim _{n \rightarrow \infty} a_{n}^{b_{n}}=a^{b}
$$

Therefore, if the limit

$$
\lim _{n \rightarrow \infty} \alpha(n) v(n)=\lim _{n \rightarrow \infty}(u(n)-1) v(n)
$$

exists, then we finally have

$$
\lim _{n \rightarrow \infty}\left[u(n)^{v(n)}\right]=e^{\lim _{n \rightarrow \infty}(u(n)-1) v(n)}
$$