## Problem Statement

Calculate the limit of the numerical sequence:

$$
\lim _{n \rightarrow \infty}\left(\frac{4 n^{2}+4 n-1}{4 n^{2}+2 n+3}\right)^{1-2 n}
$$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{4 n^{2}+4 n-1}{4 n^{2}+2 n+3}\right)^{1-2 n}=\lim _{n \rightarrow \infty}\left(\frac{4 n^{2}+2 n+3+2 n-4}{4 n^{2}+2 n+3}\right)^{1-2 n}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{2 n-4}{4 n^{2}+2 n+3}\right)^{1-2 n}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{4 n^{2}+2 n+3}{2 n-4}\right)}\right)^{1-2 n}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{4 n^{2}+2 n+3}{2 n-4}\right)}\right)^{\left(\frac{4 n^{2}+2 n+3}{2 n-4}\right)\left(\frac{2 n-4}{4 n^{2}+2 n+3}\right)(1-2 n)}= \\
& =\left(\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{4 n^{2}+2 n+3}{2 n-4}\right)}\right)^{\left(\frac{4 n^{2}+2 n+3}{2 n-4}\right)}\right)^{\lim _{n \rightarrow \infty}\left(\frac{2 n-4}{4 n^{2}+2 n+3}\right)(1-2 n)}= \\
& =\{\text { Using the second remarkable limit }\}= \\
& =e^{\lim _{n \rightarrow \infty} \frac{(2 n-4)(1-2 n)}{4 n^{2}+2 n+3}}=e^{\lim _{n \rightarrow \infty} \frac{2 n-4 n^{2}-4+8 n}{4 n^{2}+2 n+3}}=
\end{aligned}
$$

$$
\begin{aligned}
& =e^{\lim _{n \rightarrow \infty} \frac{-4 n^{2}+10 n-4}{4 n^{2}+2 n+3}}=e^{\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(-4 n^{2}+10 n-4\right)}{\frac{1}{n^{2}}\left(4 n^{2}+2 n+3\right)}}= \\
& =e^{\lim _{n \rightarrow \infty} \frac{-4+\frac{10}{n}-\frac{4}{n^{2}}}{4+\frac{2}{n}+\frac{3}{n^{2}}}}=e^{\frac{-4+0-0}{4+0+0}}=e^{-1}=\frac{1}{e}
\end{aligned}
$$

## Problem Kuznetsov Limits 7-31