Convergence of infinite product to exponential Let $x_n\to x$ be a convergent sequence of real numbers with limit $x\in\mathbb R$.
Then
$$\lim_{n\to\infty}\prod_{k=1}^n \left(1+\frac{x_k}{n}\right)=e^x.$$
Is this true? 
(Note: The formula has $n$ in the denominator and not the index $k$.)
Here is an attempt at a proof:
$$\begin{aligned}
\prod_{k=1}^n \left(1+\frac{x_k}{n}\right)&=\prod_{k=1}^n \left(1+\frac xn+\frac{x_k-x}{n}\right)\\
&=\left(1+\frac xn\right)^n+\left(1+\frac xn\right) \prod_{k=2}^n \left(\frac{x_k-x}{n}\right) +\text{stuff}\\
\end{aligned}$$
where the "stuff" will include many terms that look like a power of $\left(1+\frac xn\right)$ times some product of factors like $\left(\frac{x_k-x}{n}\right)$ and thus every other term will be decaying to zero (as $n\to\infty$) no matter how slow $x_n$ converges to $x$.
Thus taking the limit as $n\to\infty$ gives the desired result.
Now the problem is that we have infinite products still in there like $\prod_{\text{some } k} \left(\frac{x_k-x}{n}\right)$, but each factor in the infinite product has form $\frac{x_k-x}{n}$ and hence goes to zero. But I am very uncertain about this reasoning. I worry the infinite products might be very badly behaved.
I'm just wondering what technical issues I might be overlooking, and if I am making any mistakes in my argument, or if the argument is sort of correct but insufficient according to common practice.
 A: I'm not convinced. The "stuff" consists of various small terms that all will become very small, but at the same time, there will be more and more of these terms as $n \to \infty$. As with a Riemann sum, we expect to sum a lot of small quantities and take the limit, but this doesn't imply the Riemann integral is always $0$.
Here's how I would approach it:
Fix $\varepsilon > 0$, and let $N \in \Bbb{N}$ be such that
$$n \ge N \implies |x_n - x| < \varepsilon$$
and $x_n > -N$ (which we can do, as convergent sequences are bounded). Then, for $n > N$,
$$\prod_{k=1}^n \left(1 + \frac{x_k}{n}\right) = \prod_{k=1}^{N-1} \left(1 + \frac{x_k}{n}\right) \cdot \prod_{k=N}^{n} \left(1 + \frac{x_k}{n}\right),$$
where
$$\left(1 + \frac{x - \varepsilon}{n}\right)^{n-N+1} \le \prod_{k=N}^{n} \left(1 + \frac{x_k}{n}\right) \le \left(1 + \frac{x + \varepsilon}{n}\right)^{n-N+1}.$$
Since $x_n > -N$ and $n > N$, we have
$$\prod_{k=1}^{N-1} \left(1 + \frac{x_k}{n}\right) > 0,$$
so
$$\left(1 + \frac{x - \varepsilon}{n}\right)^{n-N+1} \prod_{k=1}^{N-1} \left(1 + \frac{x_n}{n}\right) \le \prod_{k=1}^{n} \left(1 + \frac{x_k}{n}\right) \le \left(1 + \frac{x + \varepsilon}{n}\right)^{n-N+1} \prod_{k=1}^{N-1} \left(1 + \frac{x_n}{n}\right).$$
Taking the limit as $n \to \infty$,
$$e^{x - \varepsilon} \le \prod_{k=1}^{\infty} \left(1 + \frac{x_k}{n}\right) \le e^{x + \varepsilon},$$
for all $\varepsilon > 0$. So, if you accept that the product converges, the limit must be $e^x$, by the continuity of $e^x$.
