Question

álgebra..................​

Answer 1

$\left(\dfrac{n+20}{n+2}\right)^{n+2017}=\left(1+\dfrac{18}{n+2}\right)^{n+2017}=\left(1+\dfrac{1}{\frac{n}{18}+\frac{1}{9}}\right)^{n+2017}=\cdots\\ \\ \\\cdots=\left[\left(1+\dfrac{1}{\frac{n}{18}+\frac{1}{9}}\right)^{\frac{n}{18}+\frac{1}{9}}\right]^{18}\left(1+\dfrac{1}{\frac{n}{18}+\frac{1}{9}}\right)^{2015}\\ \\ \\\lim\limits_{n\to+\infty}\left[\left(1+\dfrac{1}{\frac{n}{18}+\frac{1}{9}}\right)^{\frac{n}{18}+\frac{1}{9}}\right]^{18}\left(1+\dfrac{1}{\frac{n}{18}+\frac{1}{9}}\right)^{2015}=e^{18}$

Entonces

$\lim\limits_{n\to+\infty}\left(\dfrac{n+15}{n+3}\right)^{n+2017}=e^{15-3}=e^{12}\\ \\ \\\left\{\left(\dfrac{\dfrac{n+20}{n+2}}{\dfrac{n+15}{n+3}}\right)^{n+2017}\right\}\longrightarrow \dfrac{e^{18}}{e^{12}}=e^6$