Find the values of $p$ for which this sequence of functions uniformly converges. Show that $f_n\colon \mathbb{R}\to \mathbb{R}$ be defined by 
$$f_n(x) = \frac{nx}{1+n^2x^p}.$$
For $p>0$, Find the values of $p$ for which the sequence $f_n$ converges uniformly to the limit.
My attempt:
What's clear is that the sequence converges pointwise to the function $f(x)=0$.
With trial and error, I noticed that the power of $x$ should be less than the power of $n$ which is $2$. Therefore $p$ should be $1$. 
Am I thinking properly? And how to show this?
 A: We assume that $x\ge 0$, else $x^p$ is in general undefined for $x<0$.
The pointwise limit as $n\to \infty$ is zero.  We will analyze four cases to determine for which values of $p$ this convergence is uniform.  
We will see that the "problems" arise for "large" $x$ when $0<p<1$ and for "small" $x$ when $p>2$.

CASE $1$: $\displaystyle 0<p<1$
Suppose that $0<p<1$.  Take $\epsilon=\frac12$.  Then, for all $N\ge 1$ we take $x=n^{2/(1-p)}$ and any $n>N$ and find
$$\left|\frac{nx}{1+n^2x^p}\right|\ge \frac12 =\epsilon$$
This negates uniform convergence for $0<p<1$ and $x\in [0,\infty)$.

CASE $2$: $\displaystyle 2 \ge p$
Suppose that $p\ge 2$.  Take $\epsilon=1/2$.  Then for all $N\ge 1$, we take $x=n^{-2/p}$ and any $n>N$ such that 
$$\left|\frac{nx}{1+n^2x^p}\right|=\frac12 n^{1-2/p}\ge \frac12$$
And the convergence fails to be uniform for $x\in [0,\infty)$.

CASE $3$: $\displaystyle p=1$
Note the for $p=1$ we have
$$\left|\frac{nx}{1+n^2x^p}\right|\le \frac1n$$
And the convergence is uniform.

CASE $4$: $\displaystyle 1 < p<2$
For $1<p<2$, we find that $f_n'(x)=0$ at $x=\left((p-1)n^2\right)^{-1/p}$.  At that value of $x$, $f_n(x)=\frac{(p-1)^{1-1/p}}{p}n^{1-2/p}\to 0$. And convergence in the sup norm implies uniform convergence.

PUTTING IT ALL TOGETHER: 

The sequence $f_n(x)=\frac{nx}{1+n^2x^p}$ converges uniformly to $0$ for $x\in [0,\infty)$ when $1\le p<2$.  The convergence fails to be uniform for $p<1$ and fails to be uniform for $p\ge 2$.  

