In any general recurrence of the form:
$$x_n = \sum_{i=1}^k {a_i x_{n-i}}$$
you can determine an explicit formula in terms of the roots of the polynomial of degree k:
$$z^k - a_1{z^{k-1}} - a_2{z^{k-2}} - a_{k-1}z - a_k$$
If the polynomial has no repeated roots, then the form of the explicit formula will be:
$$x_n = b_1 r_1^n + b_2 r_2^n + ... + b_k r_k^n$$
Where the $b_i$ can be any numbers, and the $r_i$ are the distinct roots of the polynomial.
This means that "most of the time," $\lim\limits_{n\to \infty}{x_{n+1}/x_n}$ will be equal to the root $r_i$ with the largest absolute value such that $b_i\neq 0$. If two $r_i$ have the same largest absolute value, the convergent behavior might be odd - it possibly might never converge.
There is a lot of linear algebra involved in this - the $r_i$s are eigenvalues of a matrix which sends $(x_i,x_{i+1},...,x_{i+k-1})$ to $(x_{i+1},...,x_{i+k})$
In the case of what you call the k-nacci numbers, your polynomial is:
$$x^k-x^{k-1}-x^{k-2}-..-1 = x^k -\frac{x^k-1}{x-1}$$
I'm not sure what you can say about the roots of this polynomial when $k>2$. It's easy to show it has no repeated roots (a polynomial has no repeated roots of it is relatively prime to its derivative.)
Yes, such sequences are closely related, and the relationship does involve the golden ratio.
Let $\varphi=\frac12(1+\sqrt5)$ and $\widehat\varphi=\frac12(1-\sqrt5)$; $\varphi$ is of course the golden ratio, and $\widehat\varphi$ is its negative reciprocal. Let $a_0$ and $a_1$ be arbitrary, and define a Fibonacci-like sequence by the recurrence $a_n=a_{n-1}+a_{n-2}$ for $n\ge 2$. Then there are constants $\alpha$ and $\beta$ such that
$$a_n=\alpha\varphi^n+\beta\widehat\varphi^n\tag{1}$$
for each $n\ge 0$. Indeed, you can find them by substituting $n=0$ and $n=1$ into $(1)$ and solving the system
$$\left\{\begin{align*}
a_0&=\alpha+\beta\\
a_1&=\alpha\varphi+\beta\widehat\varphi
\end{align*}\right.$$
for $\alpha$ and $\beta$. In the case of the Fibonacci numbers themselves, $\alpha=\frac1{\sqrt5}$ and $\beta=-\frac1{\sqrt5}$; in the case of the Lucas numbers $L_n$, for which the initial values are $L_0=2$ and $L_1=1$, $\alpha=\beta=1$.
Best Answer
It is called the tribonacci series. You can find a lot of them at OEIS A000213. The long term growth is the largest root of $r^3=r^2+r+1$, which is about $1.83929$