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)?

One strategy to check for uniform convergence is to find the maximum $M_n$ of $f_n(x) = \frac{n^3x}{1 + n^4 x^2}$ over $[0,1]$ using calculus, then check whether $M_n \to 0$ (keeping in mind that $0 < f_n(x) \leq M_n$ for all $x$).
That said, the same idea works with any upper bound for $f_n(x)$ that approaches $0$ as $n\to \infty$ and does not depend on $x$. For instance, one upper bound can be attained by combining an upper bound over $[0,1/n^2]$ and another over $[1/n^2,1]$.
With the first approach we find that
$$
f_n'(x) = \frac{n^3(1 + n^4x^2) - n^3x(2n^4x)}{[1 + n^4 x^2]} = 0 \implies\\
n^3(1 + n^4 x^2)  = 2 n^3 \cdot n^4 x^2 \implies\\
(1 + n^4 x^2)  = 2 n^4 x^2 \implies\\
1 = n^4 x^2 \implies\\
x = \pm 1/n^2.
$$
We can see that the critical point at $x = 1/n^2$ is indeed a maximum since $f_n'(0) > 0$ and $f_n'(1) < 0$. Thus, our maximum is
$$
M_n = f_n(1/n^2) = \frac{n^3 \cdot \frac 1{n^2}}{1 + n^4 \cdot \frac 1{n^4}} = \frac n2.
$$
So, we find in fact that $M_n$ does not converge to $0$, which means that $f_n$ fails to converge uniformly.
