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.)
$\gcd(F_{n+1},F_{n+2}) = \gcd(F_{n+1},F_{n+1}+F_n) = \gcd(F_{n+1},F_n)$
By the induction hypothesis $\gcd(F_{n+1},F_n)=1$ so $\gcd(F_{n+1},F_{n+2})=1$
To prove $\gcd(a,b) = \gcd(a,b-a)$ use the fact that if $c\vert a$ and $c\vert b$ then $c\vert na+mb$
Best Answer
This amounts to solving the recurrence relation
$$ T_1 = t_1, \qquad T_2 = t_2, \qquad T_{n+2} = T_{n+1} + T_n. $$
By linearity, we may write $ T_n = A_n t_1 + B_n t_2 $, where $A_n$ and $B_n$ are sequences defined by the following recurrence relations.
$$ A_1 = 1, \qquad A_2 = 0, \qquad A_{n+2} = A_{n+1} + A_n, \\ B_1 = 0, \qquad B_2 = 1, \qquad B_{n+2} = B_{n+1} + B_n. $$
It is easy to check that $ A_n = F_{n-2}$ and $B_n = F_{n-1} $ solve these, where $F_n$ is the Fibonacci number starting with $F_1 = F_2 = 1$. So
$$ T_n = t_1 F_{n-2} + t_2 F_{n-1}. $$
The followings are first 10 values of $T_n$.
\begin{align*} \begin{array}{|c|c|} \hline n & T_n \\ \hline 1 & t_1 \\ 2 & t_2 \\ 3 & t_1+t_2 \\ 4 & t_1+2 t_2 \\ 5 & 2 t_1+3 t_2 \\ 6 & 3 t_1+5 t_2 \\ 7 & 5 t_1+8 t_2 \\ 8 & 8 t_1+13 t_2 \\ 9 & 13 t_1+21 t_2 \\ 10 & 21 t_1+34 t_2 \\ \hline \end{array} \end{align*}