Prove/Disprove: $ f_n(x) = \sqrt[n]{1+x^{n}} $ converges uniformly on $ [0,\infty) $.
Someone's attempt: The sequence of functions does converge uniformly.
Let $ x \in [0,1) $. $ 1 \leq (1+x^n)^{1/n} \leq (2)^{1/n} \underset{n \rightarrow \infty}{\longrightarrow} 1 $ And from sandwich, $ f_n(x) \rightarrow 1 $ for $ x \in [0,1) $.
Let $ x \in [1,\infty) $, thus $x=\sqrt[n]{x^{n}} \leqslant \sqrt[n]{1+x^{n}} \leqslant \sqrt[n]{x^{n}+x^{n}}=\sqrt[n]{2 x^{n}}=\sqrt[n]{2} \cdot x \rightarrow x$ And from sandwich, $ f_n(x) \rightarrow x $ for $ x \in [1,\infty) $.
Hence, $ f_n $ converges pointwise to $f(x)= \begin{cases}1 & x \in[0,1) \\ x & x \in[1, \infty)\end{cases}$.
For $ [ 0,1) $:
$0 \leqslant \sup \left|f_{n}(x)-f(x)\right|=\sup |\sqrt[n]{1+x}-1| \leqslant \sup |\sqrt[n]{2}-1|
\underset{n \rightarrow \infty}{\longrightarrow} \sup |1-1|=0$, hence by sandwich $\sup \left|f_{n}(x)-f(x)\right| \underset{n \rightarrow \infty}{\longrightarrow} 0$
For $[1,\infty) $:
$0 \leqslant \operatorname{sup}\left|f_{n}(x)-f(x)\right|=\sup \left|\sqrt[n]{1+x^{n}}-x\right| \leqslant \sup |\sqrt[n]{2} \cdot x-x| \underset{n \rightarrow \infty}{\longrightarrow} \sup |x-x|=0$, hence by sandwich $\sup \left|f_{n}(x)-f(x)\right| \underset{n \rightarrow \infty}{\longrightarrow} 0$.
We saw $\sup \left|f_{n}(x)-f(x)\right| \underset{n \rightarrow \infty}{\longrightarrow} 0$ on $[0,\infty) $ hence there is uniform convergence on the interval. $ \square $
Question:
Is the attempt above correct? If not then why?
I've never seen one "breaks" the $ \sup $ for several different intervals ( $[ 0,1)$ , $ [1,\infty) $ ) and then takes the limit from inside of it ( for example: $ \lim_{n \to \infty} \sup |\sqrt[n]{2}-1| = \sup | \lim_{n \to \infty} \sqrt[n]{2}-1| = \sup | 1-1| = 0 $ ).
I feel the above is very "hand" wavy and I could not justify the breaking of $\sup$ for different intervals and putting the limits inside the $ \sup $, but I would want someone elses opinion on this.
Related ( My question's about the above proof and not necessarily the solution as asked in the following questions ):
Does $f_n(x) = \sqrt[n]{1+x^n}$ converge uniformly on $[0,\infty)$?
Uniform convergence of $f_n(x) =\sqrt[n]{1+x^n}$
Checking if $f_n(x)=\sqrt[n]{1+x^n}$ uniformly converges.