Evaluate $\lim \limits_{n \to \infty\ } \Biggl( \frac{2,7}{(1+\frac{1}{n})^n}\Biggr)^n=0?$ $\lim \limits_{n \to \infty\ } \Biggl( \frac{2,7}{(1+\frac{1}{n})^n}\Biggr)^n$
I would like to replace $(1+\frac{1}{n})^n$ by $e$, and then $\frac{2,7}{e}<1$, so $\lim \limits_{n \to \infty\ } \Biggl( \frac{2,7}{(1+\frac{1}{n})^n}\Biggr)^n=0$, and I will get correct result, but I think this replacement is inadmissible.
I'm looking for the easiest way ( without advanced tools)

You're right –  one cannot replace only a part of an expression with its limit.
The simplest way consists in determining the limit of the log, using Taylor's formula at order $12$:
\begin{align}
n\log(2.7)-n^2\log\Bigl(1+\frac1n\Bigr)&=n\log(2.7)-n^2\biggl(\frac1n-\frac1{2n^2}+o\Bigl(\frac1{n^2}\Bigr)\biggr) \\
&=n\log(2.7)-n+\frac12+o(1)\\
&=n(\underbrace{\log 2.7-1}_{<\,0})+\frac12+o(1)\bigg]\to -\infty.
\end{align}
Without Taylor's formula:
From the inequalities $\;1-x<\dfrac1{1+x}<1$ $(x>0)$, you can deduce with the mean value theorem that
$$x-\frac{x^2}2<\log(1+x)<x\quad\forall x>0$$
so that
\begin{align}
n\log(2.7)-n^2\log\Bigl(1+\frac1n\Bigr)&<n\log(2.7)-n^2\biggl(\frac1n-\frac1{2n^2}\biggr) =
n(\underbrace{\log 2.7-1}_{<\,0})+\frac12
\end{align}
The same conclusion as above follows by the comparison theorem.
