The functions $f_n$ on $[0,1]$ are given by $$f_n(x) = \frac{nx}{1+n^2x^p} \ \ (p >0).$$ For what values of $p$ does the sequence converge uniformly to its pointwise limit $f$?
Consider that \begin{eqnarray*} \lim_{n \to \infty} \frac{nx}{1+n^2x^p} &=& \lim_{n \to \infty} \frac{x}{1/n + nx^p} = 0. \end{eqnarray*} We therefore have that $f_n \to 0$ pointwise. To determine what values of $p$ allow the sequence to converge such $f_n \to 0$ uniformly, we consider \begin{eqnarray*} \sup_{n \in \mathbb{N}} \left| \frac{nx}{1+n^2x^p} \right| \end{eqnarray*}
From here I've started drawing the function for different values of $p$ and $n$, note that $p$ is not necessarily a prime. However, I have made no progress, I can't see a $p$ such that $f_n \not \to f$ uniformly.
$$ f_n'(x)={n+(1-p)n^3x^{p}\over(1+n^2x^p)^2} $$ For $p\le 1$, $f'>0$ i.e. $f$ attains its maximum at $x=1$ which is ${n\over1+n^2}$. As $\lim_n{n\over1+n^2}=0$, uniform convergence occurs. For $p>1$, $f'=0$ at $x=\left({1\over n^2(p-1)}\right)^{1\over p}$ at this point $f''<0$ and therefore $f$ attains maximum here which is $$ {1\over p}(p-1)^{1-{1\over p}}n^{1-{2\over p}} $$ whose limit as $n\to\infty$ is $$ \begin{cases}\infty,& p>2\\{1\over 2},& p=2\\0,&1<p<2\end{cases} $$ So uniform convergence happens for $p\in(0,2)$.