Uniform convergence of a n-th root on $\mathbb{R}^+$ I have $f_n(x)= \sqrt[n]{n^2 + x^{4n}}$ with $x > 0$
Obliviously, It converges punctually to $f(x) = x^4$. So I have to study $$\sqrt[n]{n^2 + x^{4n}}-x^4$$
I have found that
$$\sqrt[n]{n^2 + x^{4n}}-x^4\leq \sqrt[n]{n^2} \rightarrow 1$$
But it doen't prove the non-uniform convergence.
I think that $f_n$ doesn't converge uniform because  $|\sqrt[n]{n^2 + x^{4n}}-x^4|$ doesn't have a finite $\sup_{x \in \mathbb{R}^+}$.
My intuition is correct?
 A: This is closely related to Study the pointwise convergence and uniform convergence of the sequence of functions $f_n(x)=\sqrt[n]{1+x^{2n}}$.
Since $\max(1,u)\le\left(1+u^n\right)^{1/n}\le\left(n^2+u^n\right)^{1/n}$, uniform convergence in this question implies uniform convergence in that question.
I wrote the following answer soon after this question was asked, but I was waiting for a response from the author to comments. However, since the linked question has been asked, it seems like a good time to post this answer

Assume that $u\gt0$ and $n\ge2$.
For $u^n\le n^2$,
$$
\begin{align}
\left(1+\tfrac1{\sqrt{n}}\right)^n&\ge1+\sqrt{n}\tag{1a}\\
\left(1+\tfrac1{\sqrt{n}}\right)^{4n}
&\ge\left(1+\sqrt{n}\right)^4\tag{1b}\\[6pt]
&\ge n^2\tag{1c}\\[6pt]
\left(1+\tfrac1{\sqrt{n}}\right)^4&\ge\left(n^2\right)^{1/n}\tag{1d}\\[3pt]
2^{1/n}\left(1+\tfrac1{\sqrt{n}}\right)^4&\ge\left(n^2+n^2\right)^{1/n}\tag{1e}\\[4pt]
2^{1/n}\left(1+\tfrac1{\sqrt{n}}\right)^4&\ge\left(n^2+u^n\right)^{1/n}\tag{1f}
\end{align}
$$
Explanation:
$\text{(1a)}$: Bernoulli's Inequality
$\text{(1b)}$: raise to the fourth power
$\text{(1c)}$: $\left(1+\sqrt{n}\right)^4\ge\left(\sqrt{n}\right)^4=n^2$
$\text{(1d)}$: take the $n^\text{th}$ root
$\text{(1e)}$: multiply by $2^{1/n}$
$\text{(1f)}$: $u^n\le n^2$
For $u^n\ge n^2$,
$$
\begin{align}
\left(u^n+n^2\right)^{1/n}
&=u\left(1+\frac{n^2}{u^n}\right)^{1/n}\tag{2a}\\
&\le u\left(1+\frac{n}{u^n}\right)\tag{2b}\\[3pt]
&\le u+u^{1-n/2}\tag{2c}\\[9pt]
&\le u+n^{2/n-1}\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: distributive property
$\text{(2b)}$: Bernoulli's Inequality
$\text{(2c)}$: $n\le u^{n/2}$
$\text{(2d)}$: $u\ge n^{2/n}$ and $1-n/2\le0$
Choose any $\epsilon\gt0$. Since both $2^{1/n}\left(1+\tfrac1{\sqrt{n}}\right)^4-1$ and $n^{2/n-1}$ tend to $0$ as $n\to\infty$, we can find an $N$ so that for $n\ge N$,
$$
\max\left(2^{1/n}\left(1+\tfrac1{\sqrt{n}}\right)^4-1\ ,\ n^{2/n-1}\right)\le\epsilon\tag3
$$
then $(1)$, $(2)$, and $(3)$ guarantee that for $n\ge N$,
$$
\max(1,u)\le\left(n^2+u^n\right)^{1/n}\le\max(1,u)+\epsilon\tag4
$$
Thus, setting $u=x^4$, we get that
$$
\lim_{n\to\infty}\sup_{x\ge0}\left|\left(n^2+x^{4n}\right)^{1/n}-\max\left(1,x^4\right)\right|=0\tag5
$$
That is, the convergence is uniform.
