$\lim_{n \in \mathbb{N}}n\cdot\mathrm{e}^{1/n} - n$ If I replace $\newcommand{\euler}{\mathrm{e}}$$n \to x$ where $x$ is a continuous variable. Then, I can easily apply L'Hopital's rule to $\frac{\euler^{1/x} - 1}{1/x}$ to get that the limit is $1$. 
I don't know if this is legal? Is it? 
There is a corresponding  L'Hopital's rule for sequences. But I have not had success in applying that either. 
 A: Yes, the maneuver is legal, but it's worth elaborating the relationship between limits of discrete and continuous functions extending them.
If, given a function $g : \Bbb N \to \Bbb R$ and a continuous function $\tilde g : \Bbb R^+ \to \Bbb R$ satisfying $\tilde g\vert_\Bbb N = g$, even the existence of the limit $\lim_{t \to \infty} \tilde g (t)$ is a priori possibly a stronger statement than $\lim_{n \to \infty} g(n)$.
For example, the function $\tilde g(x) = \sin 2 \pi x$ restricts to $g(n) = 0$, but $\lim_{n \to \infty} g(n) = 0$ whereas $\lim_{x \to \infty}$ does not exist. However, one can show that the converse statement holds: If $\lim_{x \to \infty} \tilde g(x)$ exists, then $\lim_{n \to \infty} g(n)$ exists and $\lim_{n \to \infty} g(n) = \lim_{x \to \infty} \tilde g(x)$.
Fortunately, this is the direction your argument uses: The function $x \mapsto x (e^{1 / x} - 1)$ on $\Bbb R^+$ restricts to the function $n \mapsto n (e^{1 / n} - 1)$ on $\Bbb N$. Since you can show that the limit of the former as $x \to \infty$ exists and compute it, the limit of the latter exists and its value coincides with it.
Remark This method relies the same underlying principle as l'Hopital's Rule, but another way to think about the problem is to expand $n (e^{1 / n} - 1)$ in a series at infinity: Since $$e^{1 / n} \sim 1 + \frac{1}{n} + O\left(\frac{1}{n^2}\right),$$
we have
$$e^{1 / n} - 1 \sim \frac{1}{n} + O\left(\frac{1}{n^2}\right),$$
and thus
$$n (e^{1 / n} - 1) \sim 1 + O\left(\frac{1}{n}\right),$$
so, just as you found using l'Hopital's Rule,
$$\lim_{n \to \infty} n (e^{1 / n} - 1) = 1 .$$
A: $\lim_{x \to 0} \frac{e^x-1}{x}=1$

We know that $\lim_{x \to x_0}f(x)=L$ if and only if ,for every sequence $x_n \to x_0$ ,we have $f(x_n) \to L$ as $n \to +\infty$

For $f(x)= \frac{e^x-1}{x}$ and $x_n=\frac{1}{n}$ and $x_0=0$ we have  that the limit is 1.
A: Since $\frac{\mathrm{d}}{\mathrm{d}x}e^x=e^x$, the Mean Value Theorem says
$$
\frac{e^{1/n}-e^0}{1/n-0}=e^{\xi}
$$
for some $\xi\in(0,1/n)$. Therefore,
$$
\begin{align}
\lim_{n\to\infty}\frac{e^{1/n}-1}{1/n}
&=\lim_{\xi\to0^+}e^{\xi}\\[6pt]
&=1
\end{align}
$$
A: Alternative approach: for any $z>0$ we have
$$ \frac{e^z-1}{z} = \frac{1}{z}\int_{0}^{z}e^t\,dt, $$
i.e. the LHS is the mean value of $e^t$ over $[0,z]$. Due to the continuity of the exponential function
$$ \lim_{z\to 0^+}\frac{e^z-1}{z} = e^0 = 1, $$
and by considering the sequence $\left\{\frac{1}{n}\right\}_{n\geq 1}$, convergent to $0$, we get
$$ \lim_{n\to +\infty} n\left(e^{1/n}-1\right) = e^0 = 1.$$
