Continuing the question from here I would like to know how to examine the uniform convergence of $\displaystyle f(x)=\frac{\ln (nx)}{n+n^4 x}, \; n \in \mathbb{N}, \; x \in (0, +\infty)$ to the $0$ function. All of the thoerems I know , fail. Definition is useless, that $\lim \sup $ test requires me to find a subsequence $a_{k_n}$ such that $\lim \sup |a_{k_n}-0| \neq 0$ which I find pretty difficult.
Therefore I don't have alternatives to attack this monster. Intuitively I know that it does not converge to the zero function uniformly but I don't have the tools to prove it.
I really need some help here!
Note that for $0 < n^3x < 1$,
$$\left|\frac{\ln(nx)}{n+ n^4x}\right|= \frac{-\ln(nx)}{n(1+ n^3x)}> \frac{-\ln(nx)}{2n}.$$
With $x_n = e^{-n}$ we have $n^3x_n < 1$ for $n$ sufficiently large, and
$$|f_n(x_n)|=\left|\frac{\ln(nx_n)}{n+ n^4x_n}\right|> \frac{-\ln(ne^{-n})}{2n}=\frac{n-\ln n }{2n}=\frac1{2}-\frac{\ln n }{2n}> \frac1{4}.$$
The last inequality is true for $n$ sufficiently large since $\ln n/n\to 0$ as $n \to \infty$.
Therefore,
$$\limsup_{n \to \infty} |f_n(x_n)| > 0,$$
and the sequence fails to converge uniformly to $0$ on $(0,\infty)$.