real analysis - Finding the limit of a sequence involving n-th roots.

We're asked to find the limit (as $n$ goes to infinity) of the following sequence: $$a_n = \frac {\sqrt[3n]{4}\ -6 \ \sqrt[3n]{2} \ + \ 9}{\sqrt[2n]{9} \ - \ 4 \ \sqrt[2n]{3}+ 4} $$.

I thought that since the limit of the numerator exists, and since the limit of the denominator also exists and is non-zero, then the limit of the fraction should be $\frac{9}{4}$.

But apperently, the limit of this sequence is $4$ as $n \rightarrow +\infty$.

I don't understand why my approach is incorrect, nor why the limit of the sequence is $4$.

Answer

It's just $$\frac{1-6+9}{1-4+4}=4$$ because for example $4^{\frac{1}{3n}}\rightarrow4^0=1$.