Find this limit: $ \lim_{n \to \infty}{(e^{\frac{1}{n}} - \frac{2}{n})^n}$ Can anyone help me with this limit:
$$ \lim_{n \to \infty}{\left(e^{\frac{1}{n}} - \frac{2}{n}\right)^n}$$
I think that I should expand $e^{\frac{1}{n}}$ as the sequence $1 + \frac{1}{n} + \frac{1}{n^{2}2!} + \frac{1}{n^{3}3!} + \dots $.
But I can't see how it helps me.
Edit: or maybe use logarithm?
 A: It's better to try computing
$$
\lim_{x\to\infty}\Bigl(e^{1/x}-\frac{2}{x}\Bigr)^{x}=
\lim_{t\to0^+}(e^t-2t)^{1/t}
$$
with the substitution $x=1/t$. Then compute the logarithm of this:
$$
\lim_{t\to0^+}\frac{\log(e^t-2t)}{t}
$$
An application of l'Hôpital's theorem will take you to the result.

 $\lim\limits_{t\to0^+}\frac{e^t-2}{e^t-2t}=-1$, so your limit is $e^{-1}$

A: Write
$$ \left(e^{1/n} - \frac{2}{n}\right)^n = e\left(1 - \frac{2}{ne^{1/n}}\right)^n$$
Let $h_n = n e^{1/n}$, note that $h_n \to \infty$ as $ n \to \infty$.
$$\left(1 - \frac{2}{ne^{1/n}}\right)^n  = \left(\left(1 - \frac{2}{h_n}\right)^{h_n}\right)^{e^{-1/n}}$$
Since $h_n \to \infty$, $\left(1 - \frac{2}{h_n}\right)^{h_n} \to e^{-2}$.
(using $\left(1 + \frac{c}{x}\right)^x \to e^c$ as $x \to \infty$)
and we also have $e^{-1/n} \to 1$.
Thus $\left(\left(1 - \frac{2}{h_n}\right)^{h_n}\right)^{e^{-1/n}} \to e^{-2}$
and so your limit is $e\times e^{-2} = e^{-1}$
A: Step 1. Preparation
\begin{align}
\lim_{n \to \infty}
{\left(e^{1/n} - \frac{2}{n}\right)^n}
=
&
\;e\cdot\lim_{n \to \infty}
{\left(1 - \frac{2}{ne^{1/n}}\right)^n}
\\
=
&
\;e\cdot\lim_{n \to \infty}
\left[{\left(1 - \frac{2}{ne^{1/n}}\right)^{ne^{1/n}}}\right]^{ \color{red}{\frac{1}{e^{1/n}}}}
\\
\end{align}
Step 2. Reduce the limit to another known limit: $\;e
\lim_{n \to \infty}{\left(1 - \frac{2}{ne^{1/n}}\right)^{ne^{1/n}}}$.
\begin{align}
\lim_{n \to \infty}
{\left(e^{1/n} - \frac{2}{n}\right)^n}
=
&
\;e\cdot\lim_{n \to \infty}
\exp\left(\log
\left[{\left(1 - \frac{2}{ne^{1/n}}\right)^{ne^{1/n}}}\right]^{\color{red}{\frac{1}{e^{1/n}}}}
\right)
\\
=
&
\;e\cdot\lim_{n \to \infty}
\exp\left(\color{red}{\frac{1}{e^{1/n}}}\log
\left[{\left(1 - \frac{2}{ne^{1/n}}\right)^{ne^{1/n}}}\right]\right)
\\
=
&
\;e\cdot
\exp\left(
\lim_{n \to \infty}\color{red}{\frac{1}{e^{1/n}}}
\cdot 
\log
\left[
\left(1 - \frac{2}{ne^{1/n}}
\right)^{ne^{1/n}}
\right]
\right)
\\
=
&
\;e\cdot
\exp\left(\color{red}{
\lim_{n \to \infty}\frac{1}{e^{1/n}}}
\cdot 
\lim_{n \to \infty}\log
\left[
\left(1 - \frac{2}{ne^{1/n}}
\right)^{ne^{1/n}}
\right]
\right)
\\
=
&
\;e\cdot
\exp\left(\color{red}{
1}
\cdot 
\lim_{n \to \infty}\log
\left[
\left(1 - \frac{2}{ne^{1/n}}
\right)^{ne^{1/n}}
\right]
\right)
\\
=
&
\;e
\lim_{n \to \infty}\exp\left(\log
\left[{\left(1 - \frac{2}{ne^{1/n}}\right)^{ne^{1/n}}}\right]\right)
\\
=
&
\;e
\lim_{n \to \infty}{\left(1 - \frac{2}{ne^{1/n}}\right)^{ne^{1/n}}}
\\
\end{align}
Step 3. Change of variable. Let $t=-ne^{1/n}$. Then
\begin{align}
\lim_{n \to \infty}
{\left(e^{1/n} - \frac{2}{n}\right)^n}
=
&
\;e
\lim_{n \to \infty}{\left(1 - \frac{2}{ne^{1/n}}\right)^{ne^{1/n}}}
\\
=
&
\;e
\lim_{t \to 0}{\left(1 + t\right)}^{\frac{-2}{t}}
\\
=
&
\;e
\lim_{t \to 0}\left[{\left(1 + t\right)}^{\frac{1}{t}}\right]^{-2}
\\
=&
e^{-1}
\\
\end{align}
