Let $x_1 = a > 0$. And $x_{n + 1} = 6 \frac{1 + x_n}{7 + x_n}$.
We know, that: $\lim_{n \to +\infty}(x_n) = x$ ($x_n$ has limit in $\mathbb{R}$).
We need to find $N(\varepsilon)$, such that:
$\forall \varepsilon > 0 \ \forall n > N(\varepsilon) \ |x_n - x| < \varepsilon$.
The main trouble I have is to compare $x_n$ and $\varepsilon$, because it's hard to get the formula, which depends from $n$.
If you look at this question, I described to steps to follow for this kind of first-order rational difference equations.
Your problem is a bit simpler and you should arrive at $$x_n=\frac{-27 (a-2) 4^n-8 (a+3) 9^n}{9 (a-2) 4^n-4 (a+3) 9^n}\tag 1$$ and a said in comments, the limit is either $-3$ or $+2$.
Since $a>0$, using $(1)$ and making $n \to \infty$, the limit is $x=2$.
Simplifying, assuming $a \neq 2$ this case would be more tha easy), we have $$|x_n-x|=\frac 5 {\frac{a+3 }{a-2}\left(\frac{9}{4}\right)^{n-1}-1 }$$ Now, you look for $$ \frac{ 5 N(\varepsilon)}{\frac{a+3 }{a-2}\left(\frac{9}{4}\right)^{n-1}-1 }< \varepsilon$$ which does not seem to be too difficult.
