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?
 A: First $n^{4n}$ is much greater than $(4n)^n$.  Informally, you can ignore the second term under the root.  Then the root is just $n^4$.  That gives a hint that we should see cancellation of the constant and $\frac 1{n^2}$ terms in the parentheses.  Indeed we can use the binomial theorem
$$\begin {align}
\left(2+\frac 1{n^2}\right)^{18}
&=2^{18}+18\cdot 2^{17}\frac 1{n^2}+{18 \choose 2}2^{16}\frac 1{n^4}+o\left(\frac 1{n^4}\right)\\
\left(4+\frac 4{n^2}\right)^9
&=4^9+9\cdot 4^8\frac 4{n^2}+{9 \choose 2}4^7\frac {4^2}{n^4}+o\left(\frac 1{n^4}\right)\\
&=2^{18}+18\cdot 2^{17}\frac 1{n^2}+{9 \choose 2}4^7\frac {4^2}{n^4}+o\left(\frac 1{n^4}\right)
\end {align}$$
and the first two terms cancel.  Evaluate the third terms, subtract them, and you will get a finite limit.
A: You may also proceed using


*

*$(\star)$: $a^n-b^n = (a-b)\sum_{k=0}^{n-1}a^kb^{n-k}$
\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*}
