I have the following limit
$$\lim_{n\rightarrow+\infty}{\left({\frac{n}{n+1}}\right)^n}$$
It leads to an indeterminate $[1^\infty]$, this means I should look into the limit of $e$. I rework the function a little bit:
$$\left({\frac{n}{n+1}}\right)^n=\left({\frac{n+1-1}{n+1}}\right)^n=\left({1-\frac{1}{n+1}}\right)^n$$
This is similar but not identical to
$$\left({1+\frac{1}{n}}\right)^n\rightarrow e$$
The major difference I notice is the $n+1$ in the denominator. Is the limit considered to approach $e^{-1}$ because for $n\rightarrow\infty$ we have $n+1\sim n$?
Best Answer
Remark that
$$\left(\frac n{n+1}\right)^n=\left(1+\frac1n\right)^{-n}$$ and by continuity of the reciprocal function, the limit of the reciprocal is the reciprocal of the limit.
This shows that in general $n+1\sim n$, unless there is some cancellation. (More specifically, $n+1-n\nsim n-n$.)