Question: > 
> Finding the maximal order in a number field 
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Suppose the number field is $K:=\mathbb Q(\alpha)$, then $\mathbb Z[\alpha]$ is not necessarily the integral closure of $\mathbb Z$ in $K$. I know the maximal order must have the smallest discriminant value among the orders in $K$. Is there a way to determine it ?

Answer: 
HINT:

$$\frac{-7^n + 8^{n-2}}{7^{n+1} + 8^{n+2}} =\frac{-(7/8)^n+8^{-2}}{7(7/8)^n+8^2}$$

Now as $-1<7/8<1,\lim\_{n\to\infty}(7/8)^n=0$
As often it's shorter with *equivalents*: as $7^n=o(8^{n-2})$ since $7<8$,
$$-7^n+8^{n-2}\sim\_{\infty}8^{n-2}$$
Similarly, $\,7^{n+1}+8^{n+2}\sim\_{\infty}8^{n+2}$, hence:
$$\frac{-7^n+8^{n-2}}{7^{n+1}+8^{n+2}}\sim\_{\infty}\frac{8^{n-2}}{8^{n+2}}=\frac 1{8^4}=\frac1{4096}.$$