Evaluating $\lim_{n\to\infty}\left(\frac{n+3}{n+1}\right)^{2n}$ I am having a hard time finding the next step on this.
Can someone tell me what is the next step and explain to me
in non-mathematical terms what I am supposed to do when I reach
this examples at this place ?
\begin{align*}
\lim_{n\to\infty}\left(\dfrac{n+3}{n+1}\right)^{2n}
&=\lim_{n\to\infty}\left(\dfrac{n+1+2}{n+1}\right)^{2n}\\
&=\lim_{n\to\infty}\left(1+\dfrac{2}{n+1}\right)^{2n}\\
&=\lim_{n\to\infty}\left(1+\dfrac{2}{n+1}\right)^{?}
\end{align*}

$$\lim_{n\to\infty}\left(\dfrac{n+3}{n+1}\right)^{2n} = \lim_{n\to\infty}\left ( \frac{n\cdot \left (1+\frac{3}{n}  \right )}{n\cdot\left (1+\frac{1}{n}  \right ) } \right )^{2n} = \lim_{n\to\infty}\left ( \frac{\left (1+\frac{3}{n}  \right )}{\left (1+\frac{1}{n}  \right ) } \right )^{2n} = \lim_{n\to\infty}\frac{\left (\left ( 1+\frac{3}{n} \right )^{n}\right)^2}{\left (\left ( 1+\frac{1}{n} \right )^{n}  \right )^{2}} =$$$$ = \frac{\left (e^{3}  \right )^{2}}{\left (e^{1}  \right )^{2}} = \frac{e^6}{e^2} = e^{6-2} = e^4.$$
