## Problem Statement

Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty}\left(\frac{10 n-3}{10 n-1}\right)^{5 n}$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{10 n-3}{10 n-1}\right)^{5 n}=\lim _{n \rightarrow \infty}\left(\frac{10 n-1}{10 n-3}\right)^{-5 n}= \\
& =\lim _{n \rightarrow \infty}\left(\frac{10 n-3+2}{10 n-3}\right)^{-5 n}=\lim _{n \rightarrow \infty}\left(1+\frac{2}{10 n-3}\right)^{-5 n}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{10 n-3}{2}\right)}\right)^{-5 n}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{10 n-3}{2}\right)}\right)^{\left(\frac{10 n-3}{2}\right)\left(\frac{2}{10 n-3}\right)(-5 n)}= \\
& =\left(\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{10 n-3}{2}\right)}\right)^{\left(\frac{10 n-3}{2}\right)}\right)^{\lim _{n \rightarrow \infty}\left(\frac{2}{10 n-3}\right)(-5 n)}= \\
& =\{\text { Using the second remarkable limit }\}= \\
& =e^{\lim _{n \rightarrow \infty}\left(\frac{2}{10 n-3}\right)(-5 n)}=e^{\lim _{n \rightarrow \infty} \frac{-10 n}{10 n-3}}= \\
& =e^{\lim _{n \rightarrow \infty} \frac{\frac{1}{n}(-10 n)}{\frac{1}{n}(10 n-3)}}=e^{\lim _{n \rightarrow \infty} \frac{-10}{10-\frac{3}{n}}}=e^{\frac{-10}{10-0}}=e^{-1}=\frac{1}{e}
\end{aligned}
$$

## Problem Kuznetsov Limits 7-20