I tried proving that the following function sequences converge uniformly but unfortunately I don't really have a clue how to do it properly and formally correct and I can't get any further. We have defined unfirom convergence as this:
$∀ \epsilon>0 \ ∃\ N \in \mathbb{N}: d_\infty (f_n, f) ≤ \epsilon \ ∀n\ ≥N$
where $d_\infty$ is the supremum-metric.
a) $$f_n=\frac{n}{n^2+x^2}$ $\ \mathbb{D} = \mathbb{R}$$ b) $$f_n=\frac{x^n}{1+x^n}$ $\ \mathbb{D} = [0, \infty)$$
I appreciate any help!
For the first one see that
$$\sup_{x\in\Bbb R}|f_n(x)|=\sup_{x\in\Bbb R}\frac{n}{n^2+x^2} =\frac1n \to 0~~~\mbox{uniformly}$$
For the second one the convergence cannot be uniform
Indeed the point-wise convergence shows that
$$ \lim_{n\to \infty} \frac{x^n}{x^n+1} = f(x)= \begin{cases}1&x> 1\\ \frac12 &x=1\\0&x> 1\end{cases}$$ the limit $f$ is clearly not continuous so the convergence cannot be uniform. it is only true on $[a,\infty)$ with $a>1$ or on $(0,b]$ with $b<1$
$$\sup_{x\ge a}|f_n(x) -1|=\sup_{x\ge a}\frac{1}{x^n+1} =\frac{1}{a^n+1} \to 0~~~\mbox{uniformly}$$
Whereas $$\sup_{x\ge 1}|f_n(x) -1|=\sup_{x\ge 1}\frac{1}{x^n+1} =\frac12 \not\to 0~~~$$