Uniform convergence of $\frac{n^3x}{1+n^4x^2}$ I have been trying to prove the uniform convergence to f = 0 on [0,1] for the sequence of functions $\frac{n^3x}{1+n^4x^2}$.
So far i have attempted to prove pointwise converge as such;
Let $ N = 1/\epsilon x  $
For every $n \ge N$ we have $   \vert\frac{n^3x}{1+n^4x^2} - 0 \vert =   \frac{n^3x}{1+n^4x^2} < \frac{1}{Nx} = \frac{1}{(1/\epsilon x)x} = \epsilon  $
Is this enough of a proof for pointwise convergence? and if so, how do i prove that the convergence is uniform (if it is in this case)?

Let $ n>0$.
For  $ x=0 $, $ f_n(0)=0$, so
$$\lim_{n\to+\infty}f_n(0)=0=f(0)$$
For $ x\in(0,1]$,
$$f_n(x)\sim \frac{n^3x}{n^4x^2}\sim \frac{1}{nx}$$
So,
$$\lim_{n\to+\infty}f_n(x)=\lim_{n\to+\infty}\frac{1}{nx}=0=f(x)$$
The convergence is pointwise to zero function at $ [0,1]$.
For the uniform convergence, let
$$g_n(x)=|f_n(x)-0|=\frac{n^3x}{1+n^4x^2}$$
the numerator of the derivative $ g'_n(x)$ is
$$n^3(1+n^4x^2)-2n^4x.n^3x=$$
$$n^3(1-n^4x^2)$$
Thus
$$M_n=\sup_{x\in[0,1]}g_n(x)=g_n(\frac{1}{n^2})=\frac n2$$
But $\lim_{n\to+\infty}M_n=+\infty$, then the convergence is Not Uniform at $[0,1]$.
