Uniform convergence of $\frac{(n+1)x}{1+(n+1)^2x^2} $ on $(-\delta , \delta)$ I want to determine whether $f_n=\displaystyle\frac{(n+1)x}{1+(n+1)^2x^2} $ converges uniformly or not on a $\delta$ neighbourhood of $0$ .
I know it converges point-wise to zero. I'm trying to prove its not uniformly convergent but I don't know where to start
 A: The convergence is not uniform even in $(0,\delta)$:
$$\left\|f_n-0\right\|=\sup_{x\in (0,\delta)}\left|\frac{(n+1)x}{1+(n+1)^2x^2}\right|=\sup_{x\in (0,\delta)}\frac{(n+1)x}{1+(n+1)^2x^2}\ge\frac{(n+1)\frac 1{n+1}}{1+(n+1)^2\frac 1{(n+1)^2}}=\frac 12$$
You might want to explain why the inequality is true (for large $n$)
A: It's a good idea to look at the graphs of the terms in the sequence.  Below is a plot of the graphs of $\color{maroon}{f_4}$,  $\color{blue}{f_7}$, $\color{darkgreen}{f_{11}}$, and $\color{pink}{f_{24}}$:

One may surmise from these plots that $(f_n)$ converges to $0$ pointwise on $(-\delta,\delta)$; but that the convergence is not uniform there, since every $f_n$ has a "peak" of height $1/2$.
And indeed, one can easily show that $(f_n)$ converges pointwise to the zero function on $(-\delta,\delta)$.  
To disprove uniform convergence, let $x_n=1/(n+1)$. These values give the peaks: $f_n(x_n)=1/2$ for every $n$, and this implies that the convergence cannot be uniform (take $\epsilon=1/2$ in the definition of uniform convergence: it is not true that $|f_n(x)|<\epsilon$ for all $x\in(-\delta,\delta)$ for any $n$).
