How can I show that a sequence is not uniform convergent? I know that
$$
f_n(x)=\frac{n^3x}{1+n^4x^3}\qquad\text{for } x\geq 1
$$
is pointwise convergent to $0$. I now need to show that it is not uniform convergent.
$f_n$ is not uniform convergent to $0$ if there is an $\epsilon_0>0$ such that for all $n\in\mathbb{N}$ I can find $x_n\in[1,\infty)$ such that $|f_n(x_n)|\geq\epsilon_0$.
In order to see what $\epsilon_0$ I need, I tried $x_n=n$ and computed
$$
|f_n(n)|=\frac{n^4}{1+n^7}>\frac{n^4}{2n^7}=\frac{1}{2n^3}
$$
but do not know what to do next!
 A: The sequence of functions you have written does in fact converge to the $0$ function uniformly. This is because, for all $n \in \mathbb{N}$ and $x \ge 1$, we have that $0 \le f_n(x) \le f_n(1)$. To see why this inequality holds, note that the derivative of $f_n$ is $\frac{n^3 - 2 n^7 x^3}{(1 + n^4 x^3)^2}$ which is less than $0$ for all $n \in \mathbb{N}$ and $x \ge 1$. Thus $f_n$ is a strictly decreasing function.
Since $\lim_{n \rightarrow \infty} f_n(1) = 0$, we have that the sequence of functions converges uniformly to the $0$ function.
A: The sequenche of functions $(f_n(x))_{n\in\mathbb{N}}$ defined by
$$f_n(x)=\frac{n x^3}{1+n^4 x^3} \ \ \ \forall x\in[1, +\infty)$$
is uniform convergent to $f(x)=0$ over $[1, +\infty)$, in fact $\forall x\in[1, +\infty)$ and $n\in\mathbb{N}\setminus\{0\}$ we have:
$$0\le |f_n(x)|=\frac{n^3 x}{1+n^4 x^3}\le \frac{n^3 x}{n^4 x^3}=\frac{1}{n x^2}\le \frac{1}{n}$$
so $$\sup_{x\in [1,+\infty)}|f_n(x)|\le\frac{1}{n}$$ and by squeeze theorem:$$0\le\lim_{n\to +\infty}\sup_{x\in[1, +\infty)}|f_n(x)|\le \lim_{n\to +\infty}\frac{1}{n}=0$$
