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?
 A: The error lies in assuming that, since $\lim_{n\to\infty}\frac n{n+2}=1$ and since $\frac1{n+2}\sim\frac1n$, then $\lim_{n\to\infty}\left(\frac n{n+2}+\frac1{n+2}\right)^n=\lim_{n\to\infty}\left(1+\frac1n\right)^n$. To see why it doesn't work, consider the limit$$\lim_{n\to\infty}n\left(\frac n{n+1}-1\right).$$It is equal to $1$, right?! However, by your argument, since $\lim_{n\to\infty}\frac{n+1}n=1$, it should be equal to$$\lim_{n\to\infty}n(1-1)=0.$$
A: $$
\lim_{n\rightarrow \infty}(\frac{n+1}{n+2})^n=\lim_{n\rightarrow \infty}\frac{1}{(\frac{n+2}{n+1})^n}=\lim_{n\rightarrow \infty}\frac{1+\frac{1}{1+n}}{(1+\frac{1}{1+n})^{n+1}}=\frac{1}{e}
$$
A: HINTS

*

*$\dfrac{n+2}{n+1}=\dfrac{(n+1)+1}{n+1}=1+\dfrac{1}{n+1}$
or

*

*$\dfrac{n+1}{n+2}=\dfrac{(n+2)-1}{n+2}=1+\dfrac{1}{-(n+2)}$
And recall that $(1+1/a_n)^{a_n}\to e$ whenever $a_n\to0$.
A: Hint:
$\dfrac{(1+1/n)^n}{(1+2/n)^n}.$
Limit of numerator and
denominator $(\not=0)$ exist.
Use: $\lim_{n \rightarrow \infty} (1+x/n)^n=e^x$, $x$ real.
A: Try replacing the variable $n$ with $x=\frac{1}{n}$: $$\lim_{n \to +\infty} (\frac{n+1}{n+2})^n = \lim_{x \to 0} (\frac{1+x}{1+2x})^\frac{1}{x} = exp(\lim_{x \to 0}\frac{1}{x} (\frac{1+x}{1+2x}-1))=exp(\lim_{x \to 0}\frac{-1}{1+2x})=exp(-1)=\frac{1}{e}$$
A: 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 $\left(1+\frac{1}{m}\right)\xrightarrow{m\rightarrow \infty}1$:
$\left[\left(1+\frac{1}{m}\right)^m\right]^{-1}\cdot\left(1+\frac{1}{m}\right)\xrightarrow{m\rightarrow \infty} \frac{1}{e}$.
