<s>[INST] Why $\lim\limits_{n \to +\infty} \bigg(\dfrac{n+1}{n+2}\bigg)^n = \frac{1}{e}$? I tried to solve this limit: $\lim\limits_{n \to +\infty} \bigg(\dfrac{n+1}{n+2}\bigg)^n$.
My approach was to re-write it as $\lim\limits_{n \to +\infty} \bigg(\dfrac{n}{n+2} + \dfrac{1}{n+2}\bigg)^n$, and since $\dfrac{n}{n+2}$ tends to 1 and $\dfrac{1}{n+2} \sim \dfrac{1}{n}$ as $n \to +\infty$, I figured the solution would be $e$, as $\lim\limits_{n \to +\infty} \bigg(1+\dfrac{1}{n}\bigg)^n = e$.
I suppose I've done something wrong, since by plotting the function I noticed the solution is $\dfrac{1}{e}$.
Where is my error?
 [/INST] You must know that $\left(1+\frac{1}{n}\right)^n \xrightarrow{n\rightarrow \infty} e$. So, we are going to play wiht it.
$\left(\frac{n+1}{n+2}\right)^n=\left[\left(\frac{n+2}{n+1}\right)^n\right]^{-1}= \left[\left(1+\frac{1}{n+1}\right)^n\right]^{-1}=\left[\left(1+\frac{1}{n+1}\right)^{(n+1)}\cdot\left(1+\frac{1}{n+1}\right)^{(-1)}\right]^{-1}=\left[\left(1+\frac{1}{n+1}\right)^{(n+1)}\right]^{-1}\cdot\left(1+\frac{1}{n+1}\right)$.
Now changing $m=n+1$ we have:
$\left[\left(1+\frac{1}{m}\right)^m\right]^{-1}\cdot\left(1+\frac{1}{m}\right)$.
Now taking limits when $n$ tends to $+\infty$, as $m=n+1$, is the same as taking limits when $m$ tends to $+\infty$. Then, as $\left(1+\frac{1}{m}\right)\xrightarrow{m\rightarrow \infty}1$:
$\left[\left(1+\frac{1}{m}\right)^m\right]^{-1}\cdot\left(1+\frac{1}{m}\right)\xrightarrow{m\rightarrow \infty} \frac{1}{e}$.
</s>