I'm given the function
\begin{equation} f_n(x)=\frac{x^n}{1+x^{2n}}, \end{equation}
where I can assume $f_n:[0,\infty)\to \mathbb{R}$. I need to determine and show the sets over which the convergence is uniform.
Clearly for $x=1$ the pointwise limit is $\frac{1}{2}$, because $f_n(1)=\frac{1}{2}$ for all $n$.
For $x \in [0,1)$, it converges pointwise to $0$, which doesn't seem too hard to show.
For $x \in (1,\infty)$ however, I believe it also converges to $0$ but I am not sure. The numerator is growing but the denominator is growing faster. Is it growing fast enough for the pointwise limit to go equal zero? Any way to see this easily?
Finally, I need to find the intervals over which this convergence is uniform. Any help here please? Thanks!
Yes you're right $f_n$ converges to $0$ for $x>1$ since in this case $$\frac{x^n}{1+x^{2n}}\sim_\infty x^{-n}$$ and since the limit function isn't continuous on $1$ then we haven't the uniform convergence on $[0,+\infty)$. However we have the uniform convergence on every interval $[0,a]$ where $0<a<1$. Indeed,
$$|f_n(x)|\le x^n\le a^n,\quad \forall x\in[0,a]$$
We have the same result on the interval $[a,+\infty)$ where $a>1$. In fact by taking the derivative of $f_n$ we can see that it's decreasing on the interval $[a,+\infty)$ so $$\sup_{x\ge a}|f_n(x)|=f_n(a)\sim_\infty a^{-n}\xrightarrow{n\to\infty}0$$