about limit of exponential function Maybe the answer is obvious. I'm sorry for this
I know for all $x \in \mathbb{R}$ that
$$
\lim_\limits{n \to \infty}\left(1 + \frac{x}{n} \right)^{n} = \exp(x).
$$
Now suppose I have a sequence $\{x_{n}\}_{n \in \mathbb{N}}$ such that
$$
\lim_\limits{n \to \infty} x_{n} = x \in \mathbb{R}.
$$
Can I also conclude that
$$
\lim_\limits{n \to \infty}\left(1 + \frac{x_{n}}{n} \right)^{n} = \exp(x)?
$$
 A: We have that $\forall \epsilon>0$ eventually 
$$\left(1 + \frac{x-\epsilon}{n} \right)^{n}\le \left(1 + \frac{x_{n}}{n} \right)^{n}\le \left(1 + \frac{x+\epsilon}{n} \right)^{n}$$
and therefore by squeeze theorem
$$e^{x-\epsilon} \le \liminf_{n \to \infty}\left(1 + \frac{x_{n}}{n} \right)^{n}\le  \limsup_{n \to \infty}\left(1 + \frac{x_{n}}{n} \right)^{n}\le e^{x+\epsilon} $$
and taking $\epsilon \to 0$ the result follows.
A: So, can one conclude that
$$n\ln\left(1+\frac{x_n}n\right)\to x?\tag{*}$$
Note that
$$n\ln\left(1+\frac{x_n}n\right)=x_n+O\left(\frac{x_n^2}{n}\right).$$
But $x_n\to x$, so $O(x_n^2/n)=O(1/n)$. Thus (*) holds, and the answer is yes.
A: By mean value theorem on $f(x)=\ln(\alpha+x)$ in $[x,x+\alpha]$ one may prove
$$\dfrac{x}{\alpha+x}<\ln(\alpha+x)-\ln(x)<1$$
this show
$$\lim_{x\to\infty}x\ln\left(1+\dfrac{\alpha}{x}\right)=1$$
which gives the result.
A: $n\log (1+x_n/n)= x_n\frac{ \log (1+x_n/n)- \log (1)}{x_n/n}=$
$x_n\log '(y_n)=x_n(1/y_n)$, where $1< y_n <1+x_n/n$.
$\lim_{n \rightarrow \infty} y_n = 1$.
Hence 
$\lim_{n \rightarrow \infty} [x_n (1/y_n)] =$ 
$[\lim_{n \rightarrow \infty}(x_n)][\lim_{ n \rightarrow \infty}(1/y_n)]= x. $
