I hope everything is going well.
I am interested in demonstrating that for all $n \in \mathbb{N}$ and $x \in B \subseteq [0,\infty)$ the following $$f_n(x) = \frac{x}{1+x^n} \text{ and } g_n(x) = \begin{cases}1,&\text{if $x \geq \frac{1}{n}$}\\nx,&\text{ if $0\leq x < \frac{1}{n}$} \end{cases}$$ converge uniformly on $B$. I am tasked with using the sup-norm for this. More specifically, I want to make use of the theorem
A sequence of functions $(f_n)_{n\in \mathbb{N}}$, with $f_n : A \to \mathbb{R}$, $n\in \mathbb{N}$, converges to $f: A \to \mathbb{R}$ uniformly on $B \subseteq A$ if $ \lim_{n\to \infty}\Vert f_n - f\Vert_{\infty;B}= 0$
Some Thoughts
I am thinking for $f_n(x)$ to consider $x \in [0,1)$ and for $g_n(x)$ to consider $x \in [0,\frac{1}{n})$. Additionally, for $f_n(x)$, I am inclined to think that since $$|f_n(x) - f(x)| = \bigg| \frac{x}{1+x^n} - 0 \bigg|< x^n < \frac{1}{n} $$ we have $$ 0 \leq \underset{x\in [0,1)}{\text{sup}}|f_n(x) -f(x)| < \frac{1}{n}$$ which along with the Squeeze Theorem could give us that $f_n(x)_{n\in \mathbb{N}}$ converges uniformly on $[0,1)$.
Any assistance would be much appreciated! Have a nice day.
Your reasoning is not correct, note that $\lim_{h \to \infty}f_h(x) = x$ for every $x \in [0,1)$ $$\Big|\frac{x}{1 + x^h} - x\Big| = \Big|\frac{-x^{h + 1}}{1 + x^h} \Big| \geq \frac{x^{h + 1}}{2}$$ Passing to the supremum we have $$\sup_{x \in [0,1)}|f_h(x) - x| \geq \frac{1}{2}$$ for every $h \in \mathbb{N}$. Therefore $(f_h)_h$ cannot converges uniformly.
For the second function I will consider $g_h : [0,1] \rightarrow \mathbb{R}$, note that each function is continuous. Let us calculate the pointwise limit: $$\lim_{h \to \infty}g_h(0) = 0$$ Note that for every $x \in (0,1]$ there exists $N \in \mathbb{N}$ such that $\frac{1}{N} \leq x$ so that $g_n(x) = 1$ for every $n \geq N$, which gives $$\lim_{h \to \infty}g_h(x) = 1$$ We have proved that the sequence $(g_h)_h$ converges pointwise to the discontinuous function $$g(x) := \begin{cases} 0 & \mbox{if } x = 0\\ 1 & \mbox{otherwise} \end{cases}$$ As a consequence the convergence cannot be uniform
Note that even if we exclude the point $0$ form the domain the convergence is still not uniform, indeed: $$|g_h(x) -1 | = \begin{cases} 0 & \mbox{if } x \geq \frac{1}{h}\\ hx & \mbox{if } x \in (0,\frac{1}{h}) \end{cases}$$ So $$\sup_{x \in (0,1]}|g_h(x) - 1| = 1$$ for every $h \in \mathbb{N}$