Weakest conditions for convergence for $x_n$ in $(1+x_n)^n$. Let $y_n$ be a sequence defined as
$$
    y_n = \left( 1 + x_n \right)^n 
$$
where $x_n$ also is some sequence. My question is: what are the loosest restrictions one can place on $x_n$ to ensure that $y_n$ converge?
My first idea was that we need $\lim_{n\to\infty} n |x_n| \leq k$, where $k \in \mathbb{R}_{\geq 0}$. However this seems like a very strict condition. 
$$ \lim_{n\to\infty} \left( 1 + \sin\frac{1}{n} \right)^n = e $$
$$ \lim_{n\to\infty} \left( 1 + \frac{1}{n^k} \right)^n = 1,\ k>1 $$
$$ \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n = e $$
 A: Hints:


*

*If we additionally assume $x_n\to 0$, then $y_n$ converges in $[-\infty, \infty]$ if and only if $n\cdot x_n$ converges in $[-\infty, \infty]$, as
$$ \log(y_n) = n \log(1+x_n) = n \cdot x_n \cdot \underbrace{\frac{1}{x_n} \int_1^{1+x_n} x^{-1} \;\mathrm d x}_{\to 1}. $$

*If $\limsup_{n\to\infty} |1+x_n| < 1$, then $y_n\to 0$.

*If $x_n$ has a limit point $p$ with $|1+p| > 1$ or $p=-2$, then $y_n$ diverges in $(-\infty, \infty)$.
You can stitch the cases together to get the characterization you want.
A: You may also consider $e^{n \log (1+x_n)}$. As the term you are interested in converges to 0, you can approximate the whole logaeithm term with its Maclaurin series. This will help you with finding the bounds on $x_n$ that ensures convergence
A: For $x\gt-1$, we have
$$
\log(1+x)\lt x\tag{1}
$$
Therefore,
$$
\log\left(1-\frac{x}{1+x}\right)\lt-\frac{x}{1+x}\tag{2}
$$
which is equivalent to
$$
\log(1+x)\gt\frac{x}{1+x}\tag{3}
$$
Putting $(1)$ and $(3)$ together, we get
$$
\frac{x}{1+x}\lt \log(1+x)\lt x\tag{4}
$$

Assume that $x_n\to0$. Since $x_n\gt-1$ eventually, we will only consider that part of the sequence.
Applying $(4)$, we get
$$
\frac{nx_n}{1+x_n}\lt n\log(1+x_n)\lt nx_n\tag{5}
$$
or equivalently
$$
n\log(1+x_n)\lt nx_n\lt(1+x_n)\,n\log(1+x_n)\tag{6}
$$
Since $1+x_n\to1$, we can take the $\liminf$ of the left inequalities and the $\limsup$ of the right inequalities in $(5)$ and $(6)$ to get
$$
\liminf_{n\to\infty}nx_n=\liminf_{n\to\infty}n\log(1+x_n)\tag{7}
$$
and
$$
\limsup_{n\to\infty}nx_n=\limsup_{n\to\infty}n\log(1+x_n)\tag{8}
$$
Therefore,

If $x_n\to0$ and either $\lim\limits_{n\to\infty}nx_n$ or $\lim\limits_{n\to\infty}n\log(1+x_n)$ exist in $[-\infty,+\infty]$, then both limits exist and are equal.

