How to calculate $\lim_{n \to \infty} {\frac {\sqrt[n]{n^{2}}-3\sqrt[n]{n}+2} {\sqrt[n]{n^{4}}+2\sqrt[n]{n^{2}}-3}}$? I am not allowed to use anything other than the following facts:
$\lim_{n \to \infty} {\frac {1}{n}}=0$
$\lim_{n \to \infty} {\sqrt[n]{n}} =1$
If $y_n \le x_n \le z_n$ and $\lim_{n \to \infty} {y_n}=\lim_{n \to \infty} {z_n}=a$ then $\lim_{n \to \infty} {x_n}=a$
I find $y_n = \frac {-\sqrt[n]{n}}{8\sqrt[n]{n^4}}$. But I don't know what is $z_n$ with $\lim z_n = -\frac {1}{8}$?
Help me find the limit of the sequence $x_n$, please