Uniform convergence of $f_n(x)=g(nx)\cdot x^{-1/6}$

Question

This is the first time I see excersise in this pattern:

Let $\displaystyle{f_n(x)=g(nx)\cdot {\frac{1}{x^{\frac{1}{6}}}}}$ $\quad g(x)={\frac{x^3+3}{x^6+x+2}}$

Is $f_n$ uniformly converge in :

(a)$(0,1]$?

(b) $[1,3]$?

(c) $[{3},\infty)$?

I just not understand what is the attitude to this kind of problem? usually I can block easly the supremum of $|f_n - f|$ or I can just find a value for $x_n$ wich indicates that it not uniform converge but here there is something different.

I can see easly that $g(x)$ getting maximum point in $[{0},\infty)$ and also that $f_n$ converge to $f(x)=0$ but nothing more than that. also, what is so special about x=3?

Answer 1

$f_n(x) \to 0$ uniformly on $[r,\infty)$ for any $r>0$. To see this note that $0 \leq f_n(x) \leq \frac 1{r^{6}} \frac {n^{3}x^{3}+3} {n^{6}x^{6}} \leq \frac 1{r^{9}}\frac 1{n^{3}}+ \frac 3 {n^{6}r^{12}}$.

On $(0,1]$ it does not converge uniformly as seen by putting $x=\frac 1 n$: note that $\sup \{f_n(s): 0<x \leq 1\}\geq f_n(\frac 1 n) \to \infty$.