Check that $$f_n(x)=\frac {x^n}{1+x^n}$$ where $x \in (0,1)$ is uniformly convergent or not?
My approach: using Mn test of uniformly convergent sequences $x_n=1-1/2n \forall n \in \mathbb{N}$ is belong to $(0,1)$
But
$$\lim_{n \to \infty}f_n(x_n) \ne 0$$ implies that $f_n(x) $ is not uniformly convergent sequence in given domain.
Is my approach correct??
Any hint and solutions is well appricated.
On
Hint. Although the pointwise limit is $f\equiv 0$, $$ \lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}\frac{x^n}{1+x^n}=0, \quad \text{for all $x\in(0,1)$}, $$ Still $$ f_n(2^{-1/n})=\frac{\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3}. $$ Hence, for $\varepsilon=1/3$, there is no $N$, such that $|f_n(x)|<\varepsilon$, for all $x\in (0,1)$.
I am just a student so I hope my answer is correct but I think that your way of thinking is correct. In all the cases here a full answer that I hope will help you or other.
-First we need to find the function towards $f_n(x)$ p.w. converges.
$\forall x \in [0;1[ \Rightarrow f_n(x) \underset{n \rightarrow \infty }{\rightarrow} 0$ and for $f_n(1)=1/2, \forall n \in \mathbb{N}$.
-Now we know that in order to prove that a sequence of function $(f_n)$ is not uniformly convergent on $I=]0;1[$ we need to find at least one sequence $(x_n) \in I$ s.t. $f_n(x_n)-f(x_n)$ doest not converge to $0$.
We choose $x_n=1-1/n \Rightarrow f(x_n)=0 $ as $0<x_n<1 \Rightarrow f(x_n) \underset{n \rightarrow \infty }{\rightarrow} 0$ and $f_n(x_n)= \frac{(1-1/n)^n}{1+(1-1/n)^n}\underset{n \rightarrow \infty }{\rightarrow}\frac{1}{1+e}$
Now by basic limit arithmetic rule we get that $(f_n(x_n)-f(x_n))\underset{n \to \infty }{\rightarrow} \frac{1}{1+e} \neq 0$.
Rem: when we use the known limit: $(1-\frac{1}{n})^n\underset{n \to \infty }{\rightarrow} \frac{1}{e}$
Q.E.D.