First, by the conclusion of # 4584, we have $\displaystyle\lim_{x\to 0^+}x^x=1$. Therefore,

$$\lim_{x\to 0^+}x^{2018x}=\lim_{x\to 0^+}\left(x^x\right)^{2018}=1$$

It then follows that

$$\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{2018x+1}dx=\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}x^{2018x+1}dx\approx\lim_{n\to\infty}n^2\int_0^{\frac{1}{n}}xdx=\boxed{\frac{1}{2}}$$