So, here is the function: $\sqrt[n]{1+x^n}$. I try to find the max of the function by deriving it: $\frac{1}{n}(1+x^n)^{\frac{1}{n}-1} \cdot nx^{n-1} = x^{n-1}(1+x^n)^{\frac{1}{n}-1}$. I tried to take natural logarithm of the function: $(n-1)\ln x + (\frac{1}{n}-1)\ln(1+x^n) \Rightarrow \ln x = -\frac{1}{n} \ln (1+x^n) \Rightarrow x = -\frac{1}{n}(1+x^n) $
I assume the way I am trying to solve it is incorrect, what could be the way to determine the uniform convergence?
First of all, if $ x $ is a real such that $ x\leq 1 $, then $ \sqrt[n]{1+x^{n}}\underset{n\to +\infty}{\longrightarrow}1 \cdot $
If $ x\geq 1 $, then $ \sqrt[n]{1+x^{n}}\underset{n\to +\infty}{\longrightarrow}x \cdot $
Thus, $ \left(f_{n}\right)_{n} $ converges (pointwise) to a function $ f $ defined us follows : \begin{aligned}f:\mathbb{R}_{+}&\rightarrow\mathbb{R}\\ x&\mapsto\left\lbrace\begin{matrix} 1&\textrm{If }x\leq 1\\ x&\textrm{If }x\geq 1\end{matrix}\right.\end{aligned}
Let $ n $ be a positive integer, we have : \begin{aligned} \left(\forall x\leq 1\right),\ \left|f\left(x\right)-f_{n}\left(x\right)\right|=\sqrt[n]{1+x^{n}}-1&=\frac{x^{n}}{\sum\limits_{k=0}^{n-1}{\left(1+x^{n}\right)^{\frac{k}{n}}}}\\&\leq\frac{1^{n}}{\sum\limits_{k=0}^{n-1}{\left(1+0^{n}\right)^{\frac{k}{n}}}}=\frac{1}{n} \end{aligned}
\begin{aligned} \left(\forall x\geq 1\right),\ \left|f\left(x\right)-f_{n}\left(x\right)\right|=\sqrt[n]{1+x^{n}}-x&=\frac{1}{\sum\limits_{k=0}^{n-1}{\left(1+x^{n}\right)^{\frac{k}{n}}x^{n-1-k}}}\\&\leq\frac{1}{\sum\limits_{k=0}^{n-1}{\left(1+0^{n}\right)^{\frac{k}{n}}1^{n-1-k}}}=\frac{1}{n} \end{aligned}
Thus, $ \left(\forall n\in\mathbb{N}\right)\left(\forall x\in\mathbb{R}_{+}\right) $, we have that : $$ \left|f\left(x\right)-f_{n}\left(x\right)\right|\leq\frac{1}{n} $$
Hence $$ \left\Vert f-f_{n}\right\Vert_{\infty}\underset{n\to +\infty}{\longrightarrow}0 $$
Which makes it possible to conclude that $ \left(f_{n}\right)_{n} $ converges $\textbf{uniformly}$ to $ f $ on $ \mathbb{R}_{+}\cdot $