5.5. Find the limit of the sequence

$$
\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2 n+3}{2 n^{2}-2 n+5}\right)^{5-7 n}
$$
Solution. In this case, we have the indeterminate form "1":

$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2 n+3}{2 n^{2}-2 n+5}\right)^{5-7 n}=\lim _{n \rightarrow \infty}\left(1+\frac{4 n-2}{2 n^{2}-2 n+5}\right)^{5-7 n}= \\
&=\lim _{n \rightarrow \infty}\left(\left(1+\frac{4 n-2}{2 n^{2}-2 n+5}\right)^{\frac{2 n^{2}-2 n+5}{4 n-2}}\right)^{\frac{(4 n-2)(5-7 n)}{2 n^{2}-2 n+5}}=\lim _{n \rightarrow \infty} e^{\frac{(4-2 / n)(5 / n-7)}{2-2 / n+5 / n^{2}}}= \\
&=e^{-14} \cong 0
\end{aligned}
$$

## CALCULATION OF THE LIMIT OF A FUNCTION