Real Analysis Limits Question

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?

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 $1$.