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


*

*$e=(1+\frac1x)^x$

*$(1+\frac{2}{n+1})^{2n}=(1+\frac{2}{n+1})^{\frac{n+1}{2}\cdot4-2}=\frac{\left((1+\frac{2}{n+1})^{\frac{n+1}{2}}\right)^4}{(1+\frac{2}{n+1})^2}$
OK, apparently some more hints are required, so here is another one:
$$\lim\limits_{x\to\infty}\frac{f(x)^p}{g(x)^q}=\frac{\left(\lim\limits_{x\to\infty}f(x)\right)^p}{\left(\lim\limits_{x\to\infty}g(x)\right)^q}$$
A: Your strategy is very good; you can make the substitution $m=n+1$ and your last limit becomes
$$
\lim_{m\to\infty}\left(\left(1+\frac{2}{m}\right)^{\!m-1}\right)^{\!2}=
\lim_{m\to\infty}
\left(\left(1+\frac{2}{m}\right)^{\!m}\right)^{\!2}
\left(1+\frac{2}{m}\right)^{\!-2}=e^4
$$
A: You can use the fact that the derivative of $e^x$ at $x = 0$ is $1,$ and the additive properties of $e^x$ to get the useful approximation $$(1+small)^{BIG} = e^{small\cdot BIG}+\cdots  $$
you have $$\left(1+\frac2{n+1} \right)^{2n} = e^{4n/(n+1)}+\cdots = e^4+\cdots $$
A: Not an efficient way, but you could also put the limit into a form where L'Hospital's Rule can be used.
\begin{align*}
&\lim_{n\to\infty}\left(\dfrac{n+3}{n+1}\right)^{2n}\\
&=\lim_{n\to\infty}(\operatorname{exp}\left(2n\ln\left(\dfrac{n+3}{n+1}\right)\right))\\
&\implies \operatorname{exp}\left(\lim_{n\to\infty}\left(2n\ln\left(\dfrac{n+3}{n+1}\right)\right)\right)\\
&=\operatorname{exp}\left(2\lim_{n\to\infty}\dfrac{\ln\left(\frac{n+3}{n+1}\right)}{\frac{1}{n}}\right)\\
&=\operatorname{exp}\left(2\lim_{n\to\infty}\dfrac{\frac{2}{n^2+4n+3}}{\frac{1}{n^2}}\right) \text{by L'Hospital's Rule}\\
&=\operatorname{exp}\left(2\lim_{n\to\infty}\dfrac{2n^2}{n^2+4n+3}\right)\\
&=\operatorname{exp}(2\cdot 2)=e^4
\end{align*}
