Real Analysis Limits Question Currently preparing for a first course in analysis exam and my answer disagrees with the model solutions but I can't see where my logic fails
The question:
Find the limit of:
$$a_n :=\frac{2^{3n}-n3^n}{n^{1729}+8^n}$$
My attempt:
First rewrite $8^n$ as $2^{3n}$
and divide through by $2^{3n}$
This gives us
$$= \frac{1-n\left(\frac{3}{8}\right)^n}{\large \frac{n^{1729}}{2^{3n}}+1}$$
and then divide through by $n$ to give
$$= \frac{\frac{1}{n}-\left(\frac{3}{8}\right)^n}{\large \frac{n^{1728}}{2^{3n}}+\frac{1}{n}}$$
My answer is zero using the following limits:
$\lim_{n\to\infty}\frac{1}{n} = 0$
and since $\frac{3}{8} <1$ $\implies\lim_{n\to\infty}\frac{3}{8}^n = 0$
The model solutions however suggest the limit is $1$, any help would be amazing.
 A: The expression $$\frac{\frac{1}{n}-(\frac{3}{8})^n}{\frac{n^{1728}}{2^{3n}}+\frac{1}{n}}$$ cannot immediatelly be used to find the limit. Sure, the numerator tends to $0$, but the denominator does too.
In fact, you should have stopped when you got to $$\frac{1-n(\frac{3}{8})^n}{\frac{n^{1729}}{2^{3n}}+1}$$
where you should notice that both the denominator and the numerator tend to $1$.
A: 
and then divide through by $n$ to give
$$= \frac{\frac{1}{n}-(\frac{3}{8})^n}{\frac{n^{1728}}{2^{3n}}+\frac{1}{n}}$$
My answer is zero using the following limits:
$\lim_{n\to\infty}\frac{1}{n} = 0$
and since $\frac{3}{8} <1$ $\implies\lim_{n\to\infty}\frac{3}{8}^n = 0$

But not only the numerator tends to zero, the denominator does too...! That would leave you with an indeterminate form, so you cannot conclude (from this) that the limit is $0$.
Instead, go one step back to the form:

$$= \frac{1-\color{blue}{n(\frac{3}{8})^n}}{\color{red}{\frac{n^{1729}}{2^{3n}}}+1}$$

and try to reason why the blue and red expressions both tend to zero, so?
