How to evaluate $\lim_{n \to \infty} \sqrt[n]{n^{(4n)}+(4n)^n}\left[\left(2+\frac{1}{n^2}\right)^{18}-\left(4+\frac{4}{n^2}\right)^9\right]$ 
$$\lim_{n \to \infty} \sqrt[n]{n^{(4n)}+(4n)^n}\left[\left(2+\frac{1}{n^2}\right)^{18}-\left(4+\frac{4}{n^2}\right)^9\right]$$

How can I deal with n-th root and the second part of the formula which I know is equal to zero?
I do not know really where to start, I tried to used $a^x = e^{x\ln(a)}$ but that doesnt seem to get me nowhere, I still have to deal with the second part which I know is equal to zero, however I feel like I should do something also with this part? 
How can I solve it, and what is the general reasoning when you see limit like this?

You may also proceed using


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*$(\star)$: $a^n-b^n = (a-b)\sum_{k=0}^{n-1}a^kb^{n-k}$
\begin{eqnarray*}  \sqrt[n]{n^{(4n)}+(4n)^n}\left[\left(2+\frac{1}{n^2}\right)^{18}-\left(4+\frac{4}{n^2}\right)^9\right]
 & =  &  \color{blue}{n^4} \sqrt[n]{1+\frac{4^n\cdot n^n}{n^{4n}}}\left[\left(  4+\frac{4}{n^2} + \frac{1}{n^4} \right)^9 - \left(4+\frac{4}{n^2}\right)^9 \right]\\
 & \stackrel{(\star)}{=}  &  \color{blue}{n^4}\sqrt[n]{1+\frac{4^n}{n^{3n}}} \frac{1}{\color{blue}{n^4} }\sum_{k=0}^8\left(4+\frac{4}{n^2} + \frac{1}{n^4} \right)^k\left(4+\frac{4}{n^2}\right)^{8-k}\\
 & = & \sqrt[n]{1+\frac{4^n}{n^{3n}}}\sum_{k=0}^8\left(4+\frac{4}{n^2} + \frac{1}{n^4} \right)^k\left(4+\frac{4}{n^2}\right)^{8-k} \\
 & \stackrel{n \to \infty}{\longrightarrow}  &  1 \cdot \sum_{k=0}^84^8 \\
 & =  & \boxed{9\cdot4^8} \\
\end{eqnarray*}
