Limit featuring 2018

We can rewrite expression as follows

$( \dfrac{1}{x^{\frac{1}{2018} } } Log(x) )^{2018}$

$( Log({ ( (x^{ \frac{1}{2018} }) } ^{ \left( \dfrac{1}{x^{\frac{1}{2018} } } \right ) } )^{2018} ) )^{2018}$

Let $x^{\frac{1}{2018} } \rightarrow \infty$

$( ( Log( 1 ))^{2018} ) )^{2018}$

$(0 )^{2018}$

$0$

Let $L_{n}=\lim_{x \to \infty} \frac {(\ln x)^{n}}{x }$ where $n$ is a nonnegative integer.

Using L'Hopital 's rule we get that $L_{n}=\lim_{x \to \infty} \frac {(\ln x)^{n}}{x }=\lim_{x \to \infty} \frac {n(\ln x)^{n-1} \times \frac {1}{x}}{1}= \lim_{x \to \infty} n\frac {(\ln x)^{n-1}}{x} = n \times L_{n-1}$ . So:

$L_{n}=n \times L_{n-1}$

Therefore $L_{n} = n! \times L_{0}$ . But $L_{0} = 0$ so $L_{n}=0$ .