uniform convergence of a real function Let's consider the sequence of functions $f_n:[0,1] \to \mathbb{R}$ , defined by $f_n(x) = \frac{nx}{1+n^2x^p}$ for $p>0$. I proved that this function converges pointwise to the constant function $f(x)=0$ . The question is about $p$ in which $p$, the convergence is uniform? 
At least I know that if $1>p>0$ then $f_n$ is increasing ( it's because $f ' _n$ >0 in that case) But I don't know how to attack the other cases 
 A: Let's compute $f_n'$ (as you allready did) to compute $f_n$'s extreme values:
\[
f_n'(x) = \frac{n(1 + n^2x^p) - n^3px^p}{(1 + n^2x^p)^2} 
\]
So we have for $p \ne 1$:
\begin{align*}
  f_n'(x) &= 0 \\
\iff n + n^3x^p - pn^3x^p &= 0\\
\iff    (1-p)n^3x^p  &= -n\\
\iff x^p &= \frac{1}{n^2(p-1)}\\
\end{align*}
For $x \in [0,1]$, this is only possible if $p > 1$. So, for $p \le 1$, $f_n$ doesn't have internal extrema, hence
\begin{align*}
  \|f_n\|_\infty &\le \max\{|f_n(0)|, |f_n(1)|\}\\
         &= |f_n(1)|\\
         &= \frac n{1+n^2} \to 0.
\end{align*}
For $p > 1$, a third value comes into play (at least for large $n$), namely 
\begin{align*}
  f_n\Bigl(\bigl(n^2(p-1)\bigr)^{-1/p}\Bigr) &= \frac{n\cdot \bigl(n^2(p-1)\bigr)^{-1/p}}{1 + n^2\bigl(n^2(p-1)\bigr)^{-1}}\\
  &= \frac{n\cdot \bigl(n^2(p-1)\bigr)^{(p-1)/p}}{n^2(p-1) + n^2}\\
  &= \frac{n^{1 + 2(p-1)/p}}{n^2}\cdot \frac{(p-1)^{(p-1)/p}}{p}\\
  &= n^{1 - 1/p} \cdot \frac{(p-1)^{(p-1)/p}}{p}
\end{align*}
Now for $n \in \mathbb N$ with $\frac 1{n^2} < p-1$ and $n^{1-1/p} \cdot \frac{(p-1)^{(p-1)/p}}{p} \ge 1$ (which finally holds for all $n$ if $p > 1$), we have
\[
\|f_n\|_\infty = n^{1 - 1/p} \cdot \frac{(p-1)^{(p-1)/p}}{p} \to \infty.
\]
