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.)
In addition to André's notes, another means of calculating solutions to these recurrence relations is to rephrase them using linear algebra as a single matrix multiply and then apply the standard algorithms for computing large powers of numbers (i.e., via binary representation of the exponent) to computing powers of the matrix; this allows for the $n$th member of the sequence to be computed with $O(\log(n))$ multiplies (of potentially exponentially-large numbers, but the multiplication can also be sped up through more complicated means).
In the Fibonacci case, this comes by forming the vector $\mathfrak{F}_n = {F_n\choose F_{n-1}}$ and recognizing that the recurrence relation can be expressed by multiplying this vector with a suitably-chosen matrix:
$$\mathfrak{F}_{n+1} = \begin{pmatrix}F_{n+1} \\\\ F_n \end{pmatrix} = \begin{pmatrix}F_n + F_{n-1} \\\\ F_n \end{pmatrix} = \begin{pmatrix} 1&1 \\\\ 1&0 \end{pmatrix} \begin{pmatrix} F_n \\\\ F_{n-1} \end{pmatrix} = M_F\mathfrak{F}_n $$
where $M_F$ is the $2\times2$ matrix $\begin{pmatrix} 1&1 \\\\ 1&0 \end{pmatrix}$. This lets us find $F_n$ by finding $M_F^n\mathfrak{F}_0$, and as I noted above the matrix power is easily computed by finding $M_F^2, M_F^4=(M_F^2)^2, \ldots$ (note that this also gives an easy way of proving the formulas for $F_{2n}$ in terms of $F_n$ and $F_{n-1}$, which are just the matrix multiplication written out explicitly; similarly, the Binet formula itself can be derived by finding the eigenvalues of the matrix $M_F$ and diagonalizing it).
Similarly, for the Tribonacci numbers the same concept applies, except that the matrix is 3x3:
$$\mathfrak{T}_{n+1} = \begin{pmatrix} T_{n+1} \\\\ T_n \\\\ T_{n-1} \end{pmatrix} = \begin{pmatrix} T_n+T_{n-1}+T_{n-2} \\\\ T_n \\\\ T_{n-1} \end{pmatrix} = \begin{pmatrix} 1&1&1 \\\\ 1&0&0 \\\\ 0&1&0 \end{pmatrix} \begin{pmatrix} T_n \\\\ T_{n-1} \\\\ T_{n-2} \end{pmatrix} = M_T\mathfrak{T}_n$$
with $M_T$ the $3\times3$ matrix that appears there; this is (probably) the most efficient all-integer means of finding $T_n$ for large values of $n$, and again it provides a convenient way of proving various properties of these numbers.
Best Answer
Let $F_n$ converge to $x$. Therefore, $$x=\frac{1}{x}+\frac{1}{x}=\frac{2}{x} $$so $$x^2=2$$providing solutions of $$x=\pm\sqrt2$$ If you try starting with negative numbers you should converge to $-\sqrt2$.